This section covers:
 Tables of Conics
 Circles
 Applications of Circles
 Parabolas
 Applications of Parabolas
 Ellipses
 Applications of Ellipses
 Hyperbolas
 Applications of Hyperbolas
 Identifying the Conic
 More Practice
Conics (circles, ellipses, parabolas, and hyperbolas) involves a set of curves that are formed by intersecting a plane and a doublenapped right cone (probably too much information!). But in case you are interested, there are four curves that can be formed, and all are used in applications of math and science:
In the Conics section, we will talk about each type of curve, how to recognize and graph them, and then go over some common applications (sorry – another way of saying “word problems”).
Always draw pictures first when working with Conics problems!
Table of Conics
Before we go into depth with each conic, here are the Conic Section Equations. Note that you may want to go through the rest of this section before coming back to this table, since it may be a little overwhelming at this point!
CONICS 
Circle Center: \(\left( {h,\,k} \right)\) 
Parabola Vertex: \(\left( {h,\,k} \right)\) 
Ellipse
Center: \(\left( {h,\,k} \right)\) \(a>b\) 
Hyperbola
Center: \(\left( {h,\,k} \right)\) \({{a}^{2}}\) before negative sign 
Equation
and Graph
(Horizontal) 
\(\displaystyle \begin{array}{c}{{\left( {xh} \right)}^{2}}+{{\left( {yk} \right)}^{2}}\\={{r}^{2}}\end{array}\) 
\(\displaystyle x=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}+h\) or \(\displaystyle xh=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\) or \(4p\left( {xh} \right)={{\left( {yk} \right)}^{2}}\) Example (positive coeff.) 
\(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\)  \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\)
Asymptotes: \(\displaystyle yk=\pm \frac{b}{a}\left( {xh} \right)\) 
Equation
and Graph
(Vertical) 
No Change 
\(\displaystyle y=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}+k\) or \(\displaystyle yk=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}\) or \(4p\left( {yk} \right)={{\left( {xh} \right)}^{2}}\) Example (positive coeff.) 
\(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\) 
\(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\)
Asymptotes: \(\displaystyle yk=\pm \frac{a}{b}\left( {xh} \right)\) 
Additional
Information 
To get \(y\): \(\begin{array}{c}y=\\\,\,\pm \sqrt{{{{r}^{2}}{{{\left( {xh} \right)}}^{2}}}}\\+k\end{array}\) 
Focal length:\(p\) Focal Width: \(4p\) Negative Coefficients: Flip parabola 
\({{c}^{2}}={{a}^{2}}{{b}^{2}}\) Length of Major Axis: \(2a\) Length of Minor Axis: \(2b\) 
\({{c}^{2}}={{a}^{2}}+{{b}^{2}}\) Length of Transverse Axis: \(2a\) Length of Conjugate Axis: \(2b\)

Note: The standard form (general equation) for any conic section is:
\(A{{x}^{2}}+Bxy+C{{y}^{2}}+Dx+Ey+F=0,\,\,\,\,\text{where}\,\,\,A,B,C,D,E,F\text{ are constants}\)
It actually turns out that if:
\({{B}^{2}}4AC<0\), if a conic exists, it is a circle or ellipse
\({{B}^{2}}4AC=0\), if a conic exists, it is a parabola
\({{B}^{2}}4AC>0\), if a conic exists, it is a hyperbola
Note: We can also write equations for circles, ellipses, and hyperbolas in terms of cos and sin, and other trigonometric functions using Parametric Equations; there are examples of these in the Introduction to Parametric Equations section.
Circles
You’ve probably studied Circles in Geometry class, or even earlier.
Circles are defined as a set of points that are equidistant (the same distance) from a certain point; this distance is called the radius of a circle.
Here is the equation for a circle, where \(r\) is the radius: \(\displaystyle \begin{array}{l}\text{Center }\left( {0,\,\,0} \right):\,\,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\\\text{Center }\left( {h,\,\,k} \right):\,\,\,\,{{\left( {xh} \right)}^{2}}+{{\left( {yk} \right)}^{2}}={{r}^{2}}\end{array}\).
If we were to solve for \(y\) in terms of \(x\) (for example, to put in the graphing calculator), we’d get: \(\displaystyle \begin{array}{l}\text{Center }\left( {0,\,\,0} \right):\,\,\,\,y=\pm \sqrt{{{{r}^{2}}{{x}^{2}}}}\\\text{Center }\left( {h,\,\,k} \right):\,\,\,\,y=\pm \sqrt{{{{r}^{2}}{{{\left( {xh} \right)}}^{2}}}}+k\end{array}\) .
Here are graphs of sample circles, with their domains and ranges:
Circle with Center \(\boldsymbol{\left( {0,0} \right)}\)  Circle with Center\(\boldsymbol{\left( {h,k} \right)}\) 
Domain: \(\left[ {2,2} \right]\) Range: \(\left[ {2,2} \right]\) 
Domain: \(\left[ {3,7} \right]\) Range: \(\left[ {8,2} \right]\) 
Sometimes we have to complete the square to get the equation for a circle. We learned how to complete the square with quadratics here in the Factoring and Completing the Square Section.
Problem:
Find the center and radius of the following circle: \({{x}^{2}}+{{y}^{2}}6x12y55=0\).
Solution:
Completing the Square for a Circle  Solution 
\(\displaystyle \begin{align}\color{#800000}{{{{x}^{2}}+{{y}^{2}}6x12y55=0}}\\{{x}^{2}}+{{y}^{2}}6x12y=55\end{align}\) \(\displaystyle \begin{align}\left( {{{x}^{2}}6x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)+\left( {{{y}^{2}}12y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=55+\,\,\underline{{\,\,\,\,\,\,}}\\\left( {{{x}^{2}}6x+\underline{{{{{\left( 3 \right)}}^{2}}}}\,} \right)+\left( {{{y}^{2}}12y+\underline{{{{{\left( 6 \right)}}^{2}}\,}}} \right)&=55+\underline{{{{{\left( 3 \right)}}^{2}}+{{{\left( 6 \right)}}^{2}}}}\\{{\left( {x\underline{3}} \right)}^{2}}+{{\left( {y\underline{6}} \right)}^{2}}&=55+9+36\\{{\left( {x3} \right)}^{2}}+{{\left( {y6} \right)}^{2}}&=100\\{{\left( {x3} \right)}^{2}}+{{\left( {y6} \right)}^{2}}&={{10}^{2}}\end{align}\) 
We move the constant to the right side, group the \(x\)’s and \(y\)’s together, and we are ready to complete the square!
Now we need to divide the coefficients (both \(x\) and \(y\)) of the middle terms by 2 and square them to complete the square. We add the squared constants to the other side.
Now we have the equation in circle form! Note that this may not necessarily work if the coefficents of \({{x}^{2}}\) and \({{y}^{2}}\) are not both positive 1 at this point, as we’ll see later.
So, for this equation, we have a circle with center \((3,6)\) and radius 10. 
You may see a “trick” problem like this, where the solution is just a point:
Problem:
Complete the square and graph: \({{x}^{2}}+{{y}^{2}}4x+2y+5=0\).
Solution:
Completing the Square for a Circle  Solution 
\(\begin{align}\color{#800000}{{{{x}^{2}}+{{y}^{2}}4x+2y+5=0}}\\{{x}^{2}}4x+{{y}^{2}}+2y=5\end{align}\) \(\begin{align}\left( {{{x}^{2}}4x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)+\left( {{{y}^{2}}+2y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=5+\,\,\underline{{\,\,\,\,\,\,}}\\\left( {{{x}^{2}}4x+\underline{{{{{\left( 2 \right)}}^{2}}}}\,} \right)+\left( {{{y}^{2}}+2y+\underline{{{{{\left( 1 \right)}}^{2}}}}\,} \right)&=5+\underline{{{{{\left( 2 \right)}}^{2}}+{{{\left( 1 \right)}}^{2}}}}\\{{\left( {x\underline{2}} \right)}^{2}}+{{\left( {y+\underline{1}} \right)}^{2}}&=5+4+1\\{{\left( {x2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}&=0\end{align}\) 
We move the constant to the right side, and we are ready to complete the square!
Now we need to divide the coefficients (both \(x\) and \(y\)) of the middle terms by 2 and square them to complete the square. We add the squared constants to the other side.
Now we have the equation in circle form, but our radius is 0!
So, for this equation, the only solution is a point at \((2,1)\) (where the center of the circle would normally be). You would just graph this point! 
Writing Equations of Circles
Sometimes you will have to come up with the equations of circle, or tangents of circles.
Problem:
Write the equation of the line that is tangent to the circle \({{\left( x3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=61\) at the point \(\left( 2,8 \right)\).
Solution:
A line tangent to a circle means that it touches the circle at one point on the outside of the circle, at a radius that is perpendicular to that line:
For this problem, since we only have one point on the tangent line \(\left( {2,8} \right)\), we’ll have to get the slope of the line of the line to get its equation. Remember here in the Coordinate System and Graphing Lines that perpendicular lines have slopes that are opposite reciprocals of each other.
Let’s draw a picture, and then get the solution:
Circle with Tangent Line  Solution 
We can get the slope of the line that connects the center of the circle \((3,2)\) and the point on the tangent line \((2,8)\), and then take the negative or opposite reciprocal to get the slope of the tangent line.
The slope of the line that contains \((3,2)\) and \((2,8)\) is \(\displaystyle \frac{{{{y}_{2}}{{y}_{1}}}}{{{{x}_{2}}{{x}_{1}}}}=\frac{{8\left( {2} \right)}}{{2\left( 3 \right)}}=\frac{{6}}{{5}}=\frac{6}{5}\). So, the slope of the tangent line is \(\displaystyle \frac{5}{6}\).
Using this slope and point \((–2,–8)\), we can use either the slopeintercept or pointslope method to get the equation; let’s use the slopeintercept: \(\displaystyle \begin{align}y&=\frac{5}{6}x+b;\,\,\,8=\frac{5}{6}\left( {2} \right)+b\,\,\,\\b&=8\frac{{10}}{6}=\frac{{29}}{3}\end{align}\)
The equation of the tangent line is \(\displaystyle y=\frac{5}{6}x\frac{{29}}{3}\).

Here’s another type of problem you might see:
Problem:
The lines \(\displaystyle y=\frac{4}{3}x\frac{5}{3}\) and \(\displaystyle y=\frac{4}{3}x\frac{{13}}{3}\) each contain diameters of a circle, and the point \(\left( {5,0} \right)\) is also on that circle. Find the equation of this circle.
Solution:
If two lines are both diameters of the same circle, where they intersect must be the center of the circle. In this case, it was easier to draw a picture to see that this is true:
So now we can get the center of the circle by finding the intersection of the two lines. Since we have another point, too, we can get the equation of the circle:
Circle Graph  Solution 
We can get the intersection of the diameters by using substitution and setting the \(y\)’s of the equations equal:
\(\displaystyle \begin{align}\frac{4}{3}x\frac{5}{3}&=\frac{4}{3}x\frac{{13}}{3}\\\frac{8}{3}x&=\frac{8}{3};\,\,\,\,x=1\\y=\frac{4}{3}\left( {1} \right)\frac{5}{3}&=3\end{align}\)
Now we know the center of the circle is \(\left( {1,3} \right)\), so the circle is in the form: \({{\left( {x+1} \right)}^{2}}+{{\left( {y+3} \right)}^{2}}={{r}^{2}}\).
To get the radius of the circle, we can use the Distance Formula \(\displaystyle \sqrt{{{{{\left( {{{x}_{2}}{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}{{y}_{1}}} \right)}}^{2}}}}\) to get the distance between the center and the given point : \(\displaystyle \sqrt{{{{{\left( {1\left( {5} \right)} \right)}}^{2}}+{{{\left( {30} \right)}}^{2}}}}=\sqrt{{25}}=5\).
The equation of circle is \({{\left( {x+1} \right)}^{2}}+{{\left( {y+3} \right)}^{2}}=25\).

Applications of Circles
Problem:
A pizza delivery area can be represented by a circle, and extends to the points \(\left( {0,18} \right)\) and \(\left( {6,8} \right)\) (these points are on the diameter of this circle). Write an equation for the circle that models this delivery area.
Solution:
If we draw a picture, we’ll see that we’ll have to use both the Distance Formula and Midpoint Formula from the Coordinate System and Graphing Lines section.
Let’s draw a picture, and then get the solution:
Circle Graph  Solution 
We first plotted the two points that form a diameter of the circle that represents the pizza delivery area: \((0,18)\) and \((6,8)\).
Since the center of a circle is midpoint between any two points of the diameter, we can use the Midpoint Theorem \(\displaystyle \left( {\frac{{{{x}_{1}}+{{x}_{2}}}}{2},\,\,\frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\) to get the center of the circle: \(\displaystyle \left( {\frac{{0+6}}{2},\frac{{18+8}}{2}} \right)=\left( {3,13} \right)\).
To get the radius of the circle, we can use the Distance Formula \(\displaystyle \sqrt{{{{{\left( {{{x}_{2}}{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}{{y}_{1}}} \right)}}^{2}}}}\) to get the distance between the center and one of the points; let’s pick \((0,18)\): \(\displaystyle \sqrt{{{{{\left( {0\left( {3} \right)} \right)}}^{2}}+{{{\left( {1813} \right)}}^{2}}}}=\sqrt{{34}}\).
The equation of circle is \({{\left( {x+3} \right)}^{2}}+{{\left( {y13} \right)}^{2}}=34\).

Parabolas
Let’s revisit parabolas (a type of quadratic), but go into a little more depth here.
We studied Parabolas in the Introduction to Quadratics section, but we only looked at “vertical” parabolas (that either go up or down); a parabola with negative coefficient faces down (“cup down”).
We remember that a parabola is in the form \(y=a{{\left( xh \right)}^{2}}+k\), where \(\left( {h,\,k} \right)\) is the vertex and \(x=h\) is the axis of symmetry or line of symmetry (LOS); this is a “vertical” parabola. Note that this can also be written \(yk=a{{\left( {xh} \right)}^{2}}\) or \(b\left( {yk} \right)={{\left( {xh} \right)}^{2}}+k\), where \(\displaystyle b=\frac{1}{a}\).
Parabolas can also be in the form \(x=a{{\left( yk \right)}^{2}}+h\), where \(\left( {h,\,k} \right)\) is the vertex, and \(y=k\) is the LOS; this is a “horizontal” parabola. The line of symmetry (LOS) is a line that divides the parabola into two parts that are mirror images of each other. In these cases, parabolas with a negative coefficient faces left.
Technically, a parabola is the set of points that are equidistant from a line (called the directrix) and another point not on that line (called the focus, or focal point). For vertical (up and down) parabolas, the directrix is a horizontal line (“\(y=\)”), and for horizontal (sideways), the directrix is a vertical line (“\(x=\)”).
If \(p\) is the distance from the vertex to the focus point (called the focal length), it is also the distance from the vertex to the directrix. This makes the distance from the focus to the directrix is \(2p\). Note that the focus is always “inside” the parabola on the line of symmetry, and the directrix is “outside” the parabola.
Also note that the line perpendicular to the line of symmetry (and thus parallel to the directrix) that connects the focus to the sides of the parabola is called the latus chord, latus rectum, or focal width, focal diameter, focal chord or focal rectum; the length of this chord is \(4p\).
So to draw the parabola, if you know \(p\), you can just go out \(2p\) on either side of the focus to get more points!
Here is a parabola with center \(\left( {0,\,0} \right)\):
If the vertex is at the origin \(\left( {0,\,0} \right)\), the equation of the parabola is \(y=a{{x}^{2}}\), and \(\displaystyle a=\frac{1}{{4p}}\); if you do the algebra, it follows that that \(\displaystyle p=\frac{1}{4a}\).
For example, if \(p=4\) (length of focus to vertex), the equation of the parabola would be \(\displaystyle y=\frac{1}{{4\left( 4 \right)}}{{x}^{2}}\,=\frac{1}{{16}}{{x}^{2}}\).
Here are the four different “directions” of parabolas and the generalized equations for each. It looks complicated, but it’s really not that bad; just remember to draw the parabolas, and you’ll get the hang of it pretty quickly.
Also remember that the “\(h\)” always goes with the “\(x\)” and the “\(k\)” always goes with the “\(y\)”.
Vertical Parabola  Horizontal Parabola 
Positive Coefficient At \(\left( {0,0} \right):\,\,\,\,\,y=a{{x}^{2}}\) \(y=a{{\left( {xh}\right)}^{2}}+k\) or \(yk=a{{\left( {xh} \right)}^{2}}\) \(\displaystyle y=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}+k\) or \(\displaystyle yk=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}\) or \(4p\left( {yk} \right)={{\left( {xh} \right)}^{2}}\) Vertex: \(\left( {h,\,\,k} \right)\) Axis of Symmetry: \(x=h\) 
Positive Coefficient At \(\left( {0,0} \right):\,\,\,\,\,x=a{{y}^{2}}\) \(x=a{{\left( {yk} \right)}^{2}}+h\) or \(xh=a{{\left( {yk} \right)}^{2}}\) \(\displaystyle x=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}+h\) or \(\displaystyle xh=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\) or \(4p\left( {xh} \right)={{\left( {yk} \right)}^{2}}\) Vertex: \(\left( {h,\,\,k} \right)\) Axis of Symmetry: \(y=k\) 
Negative Coefficient At \(\left( {0,0} \right):\,\,\,\,y=a{{x}^{2}}\) \(y=a{{\left( {xh} \right)}^{2}}+k\) or \(yk=a{{\left( {xh} \right)}^{2}}\) \(\displaystyle y=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}+k\) or \(\displaystyle yk=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}\) or \(4p\left( {yk} \right)={{\left( {xh} \right)}^{2}}\) Vertex: \(\left( {h,\,\,k} \right)\) Axis of Symmetry: \(x=h\) 
Negative Coefficient At \(\left( {0,0} \right):\,\,\,\,\,x=a{{y}^{2}}\) \(x=a{{\left( {yk} \right)}^{2}}+h\) or \(xh=a{{\left( {yk} \right)}^{2}}\) \(\displaystyle x=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}+h\) or \(\displaystyle xh=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\) or \(4p\left( {xh} \right)={{\left( {yk} \right)}^{2}}\) Vertex: \(\left( {h,\,\,k} \right)\) Axis of Symmetry: \(y=k\) 
Note that sometimes (as in the problem below) we have to complete the square to get the equation in parabolic form; we did this here in the Solving Quadratics by Factoring and Completing the Square Section.
Let’s do some problems!
Problem:
Identify the vertex, axis of symmetry, focus, equation of the directrix, and domain and range for the following parabolas, then graph the parabola: (a) \(y4=2{{\left( x3 \right)}^{2}}\) (b) \(\displaystyle {{y}^{2}}4y+2x8=0\). (This is in standard or general form).
Solution:
It’s typically easier to graph the parabola first, and then answer the questions.
Parabola Graph  Solution 
\(\displaystyle y4=\frac{1}{{16}}{{\left( {x3} \right)}^{2}}\)
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left[ {4,\infty } \right)\) 
We see the equation is in the form \(\displaystyle yk=\frac{1}{{4p}}{{\left( {xh} \right)}^{2}}\), where \(p\) is the focal length. Thus, the vertex is \(\left( {3,4} \right)\), and the axis/line of symmetry (LOS) is \(x=3\).
Since \(4p=16\), the focal length is 4.Since the focus point is “inside” the parabola, it is up 4 from \(\left( {3,4} \right)\), so it is \(\left( {3,8} \right)\). The directrix is “outside” the parabola the same distance away from the vertex, so it is at \(y=0\).
To complete the graph, we can use the fact that the latus chord (line perpendicular to the LOS through the focus to either side of the parabola) is \(4p\), so we can go over \(2p\) (8) from each side of the focus to get points on the parabola. (We could also plug random points in the equation for \(x\) to get \(y\)). 
\({{y}^{2}}4y+2x8=0\) \begin{align}x4&=\frac{1}{2}\left( {{{y}^{2}}4y} \right)\\x4\color{#117A65}
Domain: \(\left( {\infty ,6} \right]\) Range: \(\left( {\infty ,\infty } \right)\) 
Since we have an \(x\) and \({{y}^{2}}\), let’s try to put the standard equation in the form \(\displaystyle xh=\left( – \right)\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\), where \(p\) is the focal length.
When we solve for \(x\) (getting \(x\) and a constant on one side), we’ll see that we need to complete the square so we can get the equation in parabola form. We get \(x6=.5{{\left( {y2} \right)}^{2}}\).
We can see from the equation of a parabola that it is a “horizontal” parabola that opens up to the left with vertex \(\left( {6,2} \right)\) and axis/line of symmetry (LOS) \(y=2\).
Since \(4p=2\), the focal length is \(\displaystyle \frac{1}{2}\). Since the focus point is “inside” the parabola, it is to the left \(\displaystyle \frac{1}{2}\) from \((6,2)\), so it is \((5.5,2)\). The directrix is “outside” the parabola the same distance away from the vertex, so it is \(x=6.5\).
To complete the graph, we can use the fact that the latus chord (line perpendicular to the LOS through the focus to either side of the parabola) is \(4p\), so we can go over \(2p\) (1) from each side of the focus to get points on the parabola. (We could also plug random points in the equation for \(x\) to get \(y\)).

Writing Equations of Parabolas
Problems:
 Write the equation of a parabola with a vertex of \(\left( {2,4} \right)\) and focus point \(\left( {0,4} \right)\). Also find the domain and range of the parabola.
 Write the equation of a parabola with a focus at \(\left( {2,4} \right)\) and a directrix of \(y=9\). Also find the domain and range of the parabola.
Solutions:
Parabola Graph 
Solution 
Domain: \(\left[ {2,\infty } \right)\) Range: \(\left( {\infty ,\infty } \right)\) 
vertex of \(\left( {2,4} \right)\) and focus point \(\left( {0,4} \right)\)
It’s best to first plot the points, so we can see the direction of the parabola. We can see that it’s a horizontal parabola that opens up to the right, since the focus is inside the graph.
We know that the parabola is in the form \(\displaystyle xh=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\), where \(p\) is focal length.
Since the focal length (length from the vertex to the focus) is 2, and the vertex is \(\left( {2,4} \right)\), we have \(\displaystyle x\left( {2} \right)=\frac{1}{{4\left( 2 \right)}}{{\left( {y4} \right)}^{2}}\), or \(\displaystyle x+2=\frac{1}{8}{{\left( {y4} \right)}^{2}}\). You can also write the parabola as \(\displaystyle x=\frac{1}{8}{{\left( {y4} \right)}^{2}}2\) or \(8\left( {x+2} \right)={{\left( {y4} \right)}^{2}}\). 
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,6.5} \right)\) 
focus at \(\left( {2,4} \right)\) and a directrix of \(y=9\) It’s best to first plot the points, so we can see the direction of the parabola. We can see that it’s a vertical parabola that opens down, since the since the directrix is horizontal and the focus is below it.
We know that the parabola is in the form \(\displaystyle yh=\frac{1}{{4p}}{{\left( {xk} \right)}^{2}}\), where \(p\) is focal length.
Since the length from the focus to the directrix is \(5\,\,\,(94)\), and the vertex is exactly in between the focus and directrix, the focal length (length from the vertex to the focus) is \(\displaystyle \frac{5}{2}=2.5\). The vertex is \(\left( {2,6.5} \right)\).
We have \(\displaystyle y6.5=\frac{1}{{4\left( {2.5} \right)}}{{\left( {x\left( {2} \right)} \right)}^{2}}\), or \(\displaystyle y6.5=\frac{1}{{10}}{{\left( {x+2} \right)}^{2}}\). This is also \(\displaystyle y=\frac{1}{{10}}{{\left( {x+2} \right)}^{2}}+6.5\) or \(10\left( {y6.5} \right)={{\left( {x+2} \right)}^{2}}\).

Here’s another parabola problem that’s a bit tricky:
Problem:
Write the equation and graph a parabola with focus \(\left( {2,7} \right)\) that opens to the right, and contains the point \(\left( {6,1} \right)\).
Solution:
Parabola Graph  Solution 
Parabola with Focus \(\left( {2,7} \right)\) that opens to the right, parabola contains the point \(\left( {6,1} \right)\)

We know that the equation for a parabola that opens up to the right is \(\displaystyle xh=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}\), where \(p\) is the focal point. By drawing the parabola, we can see that the vertex will be \(p\) units to the left of the focus, \(\left( {2,7} \right)\), so the vertex will be at \(\left( {2p,7} \right)\). From this, we know that \(h=2p\) and \(k=7\).
We can now plug in the point \(\left( {6,1} \right)\) for \(x\) and \(y\), and \(\left( {2p,7} \right)\) for \(\left( {h,k} \right)\), and solve for \(p\): \(\require{cancel} \begin{array}{c}6\left( {2p} \right)=\frac{1}{{4p}}{{\left( {\left( {1} \right)\left( {7} \right)} \right)}^{2}}\\4p\left( {8+p} \right)={{6}^{2}};\,\,\,\,32p+4{{p}^{2}}=36\\{{p}^{2}}+8p9=0\,\,\\\,\,\left( {p+9} \right)\left( {p1} \right)=0\,\\p=\cancel{{9}},\,1\end{array}\)
Since \(p=1\), the vertex is at \(\left( {21,7} \right)\), or \(\left( {3,7} \right)\).
The equation for the parabola is: \(\displaystyle x+3=\frac{1}{4}{{\left( {y+7} \right)}^{2}}\).

Applications of Parabolas
The main application of parabolas, like ellipses and hyperbolas, are their reflective properties (lines parallel to the axis of symmetry reflect to the focus). They are very useful in realworld applications like telescopes, headlights, flashlights, and so on.
Problem:
The equation \(\displaystyle \frac{1}{32}{{x}^{2}}\) models cross sections of parabolic mirrors that are used for solar energy. There is a heating tube located at the focus of each parabola; how high is this tube located above the vertex of its parabola?
Solution:
For problems like these, unless otherwise noted, just assume the vertex of the parabola is at \(\left( {0,0} \right)\). Since we know that the equation of a parabola is \(y=a{{x}^{2}}\), where \(\displaystyle a=\frac{1}{4p}\) and \(\displaystyle p=\frac{1}{4a}\), then for \(\displaystyle \frac{1}{32}{{x}^{2}}\), we have \(\displaystyle \frac{1}{32}=\frac{1}{4p}\). We can either cross multiply or just do mental math to see that \(p=8\). The heating tube needs to be located 8 units above the vertex of the parabola.
Problem:
A searchlight has a parabolic reflector (has a cross section that forms a “bowl”). The parabolic “bowl” is 16 inches wide from rim to rim and 12 inches deep. The filament of the light bulb is located at the focus. (a) What is the equation of the parabola used for the reflector? (b) How far from the vertex is the filament of the light bulb?
Solution:
Let’s graph this particular parabola, again putting the vertex at \(\left( {0,0} \right)\).
Graph of Parabola  Solution 
(a) Since the vertex is at \((0,0)\), we can put the parabola in the form \(y=a{{x}^{2}}\), where \(\displaystyle a=\frac{1}{{4p}}\), and \(p\) is focal length.
It’s best to draw the parabola, and since the diameter of the bowl is 16 and the height is 12, we know that the point \((8,12)\) is on the graph (we have to divide the diameter by 2, since that is the distance all the way across).
We can find \(a\), using the equation \(y=a{{x}^{2}}\), where \(x=8\) and \(y=12\). So \(12=a{{\left( 8 \right)}^{2}}\), and we see that \(\displaystyle a=\frac{{12}}{{64}}=.1875\). The equation of the parabola then is \(y=.1875{{x}^{2}}\).
(b) To find how far the filament is, we need to find the focus. Since \(\displaystyle p=\frac{1}{{4a}}\), we have \(\displaystyle p=\frac{1}{{4\left( {.1875} \right)}}=\frac{1}{{.75}}=\frac{4}{3}\). Therefore, the distance from the vertex to the filament (focus) is \(\displaystyle \frac{4}{3}\) inch.

Problem:
The cables of the middle part of a suspension bridge are in the form of a parabola, and the towers supporting the cable are 600 feet apart and 100 feet high. What is the height of the cable at a point 150 feet from the center of the bridge?
Solution:
Let’s draw a picture of the bridge, and place the middle of the cable (vertex) at the point \(\left( {0,0} \right)\).
Graph of Parabola  Solution 
We know the distance between the towers is 600 feet and they are 100 feet tall. Therefore, we can place a point at \((300,100)\) at the top of a tower (since the bridge is symmetrical).
The problem asks for the height of the parabola 150 feet from the center, so we need the \(y\) value when the \(x\) value is 150.
We can get the equation of the parabola with \(y=a{{x}^{2}}\), and plug in the point \((300,100)\) to get the \(a\) value: \(100=a{{\left( {300} \right)}^{2}}\); \(\displaystyle a=\frac{{100}}{{90000}}=\frac{1}{{900}}\). The equation of the parabola then is \(\displaystyle y=\frac{1}{{900}}{{x}^{2}}\).
To find the \(y\) value when \(x=150\), plug in \(x\): \(\displaystyle y=\frac{1}{{900}}{{\left( {150} \right)}^{2}}=25\). Therefore, the height of the cable 150 feet from the center of the bridge is 25 feet.

Ellipses
An ellipse sort of looks like an oval or a football, and is the set of points whose distances from two fixed points (called the foci) inside the ellipse is always the same, \({{d}_{1}}+{{d}_{2}}=2a\). The distance \(2a\) is called the constant sum or focal constant, and \(a\) is the distance between the center of the ellipse to a vertex (you usually don’t have to worry about the \({{d}_{1}}\) and \({{d}_{2}}\)). \(a\) (the length of the center to the vertices) is always bigger than \(b\) (the length of the center to the covertices).
The equation of a “horizontal” ellipse that is centered on the origin \(\left( {0,0} \right)\) is \(\displaystyle \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\) (what’s under the \(x\) is larger than what’s under the \(y\)). The equation of a transformed horizontal ellipse with center \((h,k)\) is \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\).
The length of the longest axis (called the major axis) is always \(2a\), and this is along the \(x\)axis for a horizontal ellipse. Again, the distance from the center of the ellipse to a vertex is \(a\), so the vertices are at \(\left( \pm a,\,\,0 \right)\).
The length of the smaller axis (called the minor axis or) is \(2b\), and this is along the \(y\)axis for a horizontal ellipse. Again, the distance from the center of the ellipse to a covertex is \(b\), so the covertices are at \(\left( 0,\,\,\pm b \right)\).
The focuses or foci always lie inside the ellipse on the major axis, and the distance from the center to a focus is \(c\). The foci are at \(\left( \pm c,\,\,0 \right)\) for this type of ellipse, and it turns out that \({{a}^{2}}{{b}^{2}}={{c}^{2}}\).
Note that a circle happens when \(a\) and \(b\) are the same in an ellipse, so a circle is a special type of ellipse, but for all practical purposes, circles are different than ellipses. Sometimes you will be asked to get the eccentricity of an ellipse \(\displaystyle \frac{c}{a}\), which is a measure of how close to a circle the ellipse is; when it is a circle, the eccentricity is 0. Also, the area of an ellipse is \(\pi ab\).
Note also that the focal width (focal chord, or focal rectum) of an ellipse is \(\displaystyle \frac{{2{{b}^{2}}}}{a}\); this the distance perpendicular to the major axis that goes through the focus.
Here are the two different “directions” of ellipses and the generalized equations for each:
Horizontal Ellipse  Vertical Ellipse 
At \(\displaystyle \left( {0,0} \right):\,\,\,\,\frac{{{{x}^{2}}}}{{{{a}^{2}}}}+\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1\) General: \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\) \({{a}^{2}}{{b}^{2}}={{c}^{2}}\) Center: \(\left( {h,\,k} \right)\) Foci: \(\left( {h\pm c,\,k} \right)\) Vertices: \(\left( {h\pm a,\,k} \right)\) CoVertices: \(\left( {h,\,k\pm b} \right)\) 
At \(\displaystyle \left( {0,0} \right):\,\,\,\,\frac{{{{x}^{2}}}}{{{{b}^{2}}}}+\frac{{{{y}^{2}}}}{{{{a}^{2}}}}=1\) General: \(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\) \({{a}^{2}}{{b}^{2}}={{c}^{2}}\) Center: \(\left( {h,\,k} \right)\) Foci: \(\left( {h,\,k\pm c} \right)\) Vertices: \(\displaystyle \left( {h,\,k\pm a} \right)\) CoVertices: \(\left( {h\pm b,\,k} \right)\)

Major Axis Length \(=2a\) Minor Axis Length \(=2b\) Focal Width \(\displaystyle =\frac{{2{{b}^{2}}}}{a}\) 
You also may have to complete the square to be able to graph an ellipse, like we did here for a circle. (And since you always have to have a “1” after the equal sign, you may have to divide all terms by the constant on the right, if it isn’t “1”).
Let’s put it all together and graph some ellipses:
Problem:
Identify the vertices, covertices, foci, and domain and range for the following ellipses; then graph: (a) \(9{{x}^{2}}+49{{y}^{2}}=441\) (b) \(\displaystyle \frac{{{\left( x+3 \right)}^{2}}}{4}+\frac{{{\left( y2 \right)}^{2}}}{36}=1\).
Solution:
It’s typically easier to graph the ellipse first, and then answer the questions:
Ellipse  Math/Notes 
(a) \(9{{x}^{2}}+49{{y}^{2}}=441\)
Domain: \(\left[ {7,\,7} \right]\) Range: \(\left[ {3,\,3} \right]\) 
We first need to get our equation into the form of an ellipse by dividing all terms by 441: \(\displaystyle \frac{{9{{x}^{2}}}}{{441}}+\frac{{49{{y}^{2}}}}{{441}}=\frac{{441}}{{441}}\), or \(\displaystyle \frac{{{{x}^{2}}}}{{49}}+\frac{{{{y}^{2}}}}{9}=1\). We will use the equation \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since \(49>9\) (horizontal ellipse).
This would make \({{a}^{2}}=49\), so \(a=7\). Since the ellipse’s center is \((0,0)\), the vertices are \((7,0)\) and \((7,0)\). \({{b}^{2}}=9\), so \(b=3\); the covertices are \((0,3)\) and \((0,3)\).
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}{{b}^{2}}=499=40\). \(c=\sqrt{{40}}\text{ }\left( {\text{or }2\sqrt{{10}}} \right)\), and the foci are \(\displaystyle \left( {\pm \sqrt{{40}},0} \right)\). These can be difficult to graph, but just estimate \(\sqrt{{40}}\) to be close to 6.
Notice that the vertices and foci lie along the horizontal line \(y=0\). The length of the major axis is \(2a=14\) and the length of the minor axis is \(2b=6\). 
(b) \(\displaystyle \frac{{{{{\left( {x+3} \right)}}^{2}}}}{4}+\frac{{{{{\left( {y2} \right)}}^{2}}}}{{36}}=1\)
Domain: \(\left[ {5,1} \right]\) Range: \(\left[ {4,\,\,8} \right]\) 
We will use equation: \(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since \(36>4\) (vertical ellipse). We see that the center of the ellipse is at \((3,2)\), so we can first plot that point.
\({{a}^{2}}=36\), so \(a=6\). Since the center is \((–3 ,2)\), the vertices are \((3,26)\) and \((3,2+6)\), or \((3,4)\) and \((3,8)\). \({{b}^{2}}=4\), so \(b=2\); the covertices are \((32,2)\) and \((3+2,2)\) or \((5,2)\) and \((1,2)\).
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}{{b}^{2}}=364=32\). \(c=\sqrt{{32}}\text{ }\left( {\text{or }4\sqrt{2}} \right)\), and the foci are \(\left( {3,\,\,2\pm \sqrt{{32}}} \right)\). These can be difficult to graph, but just estimate the point: for example, \(2\,+\sqrt{{32}}\) is about \(2+5.5\), or about 7.5.
Notice that the vertices and foci are along the vertical line \(x=3\). The length of the major axis is \(2a=12\) and the length of the minor axis is \(2b=4\). 
Here’s one where you have to Complete the Square to be able to graph the ellipse:
Problem:
Identify the vertices, covertices, foci and domain and range for the following ellipse; then graph: \(4{{x}^{2}}+{{y}^{2}}+24x+2y=33\)
Solution:
Ellipse  Math/Notes 
\(4{{x}^{2}}+{{y}^{2}}+24x+2y=33\)
\(\displaystyle \begin{array}{c}4{{x}^{2}}+24x+{{y}^{2}}+2y=33\\4\left( {{{x}^{2}}+6x} \right)+{{y}^{2}}+2y=33\\4\left( {{{x}^{2}}+6x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)+\left( {{{y}^{2}}+2y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)=33+\,\,\underline{{\,\,\,\,\,\,}}\\4\left( {{{x}^{2}}+6x+\color{#2E8B57}{{\underline{{{{{\left( 3 \right)}}^{2}}}}}}} \right)+\left( {{{y}^{2}}+2y+\color{#2E8B57}{{\underline{{{{{\left( 1 \right)}}^{2}}}}}}\,} \right)=33+\color{#2E8B57}{{\underline{{4{{{\left( 3 \right)}}^{2}}+{{{\left( 1 \right)}}^{2}}}}}}\\4{{\left( {x+\underline{3}} \right)}^{2}}+{{\left( {y+\underline{1}} \right)}^{2}}=33+36+1\\4{{\left( {x+3} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}=4\end{array}\) \(\displaystyle \frac{{{{{\left( {x+3} \right)}}^{2}}}}{1}+\frac{{{{{\left( {y+1} \right)}}^{2}}}}{4}=1\)
Domain: \(\left[ {4,\,\,2} \right]\) Range: \(\left[ {3,\,\,1} \right]\) 
We need to first complete the square so we can get the equation in ellipse form. Put all the \(x\)’s and \(y\)’s together with the constant term on the other side. There can’t be a number (coefficient) before the \({{x}^{2}}\) or \({{y}^{2}}\), so factor out the 4 in front of the \({{x}^{2}}\). Don’t forgot to include it when adding to the right side.
After you complete the square, divide all terms by 4, so we have a 1 on the right.
We will use equation: \(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since \(4>1\) (vertical ellipse).
We see that the center of the ellipse is at \((–3,–1)\), so we can first plot that point.
\({{a}^{2}}=4\), so \(a=2\). Since the center is \((–3,–1)\), the vertices are \((–3,–1–2)\) and \((–3,–1+2)\), or \((–3,–3)\) and \((–3,1)\). \({{b}^{2}}=1\), so \(b=1\). Thus, the covertices are \((31,1)\) and \((–3+1,–1)\), or \((–4,–1)\) and \((–2,–1)\).
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}{{b}^{2}}=41=3\). \(c=\sqrt{3}\), and the foci are \(\displaystyle \left( {3,1\,\pm \sqrt{3}} \right)\).
Notice that the vertices and foci are along the vertical line \(x=3\). The major axis is \(2a=4\), and the minor axis is \(2b=2\). 
Writing Equations of Ellipses
You may be asked to write an equation from either a graph or a description of an ellipse:
Problem
Write the equation of the ellipse:
Solution:
We can see that the ellipse is 10 across (the major axis length) and 4 down (the minor axis length). So, \(2a=10\), and \(2b=4\). We can also see that the center of the ellipse \(\left( {h,k} \right)\) is at \(\left( {4,3} \right)\).
Since the ellipse is horizontal, we’ll use the equation \(\displaystyle \frac{{{\left( xh \right)}^{2}}}{{{a}^{2}}}+\frac{{{\left( yk \right)}^{2}}}{{{b}^{2}}}=1\). Plug in our values for \(a,\,b,\,h,\,\,\text{and}\,k\), and we get \(\displaystyle \frac{{{\left( x4 \right)}^{2}}}{25}+\frac{{{\left( y+3 \right)}^{2}}}{4}=1\).
Note that we didn’t have to have the coordinates of the foci to obtain the equation of the ellipse.
Problem:
Find the equation of this ellipse, graph, and find the domain and range: Endpoints of major or minor axis at \(\left( {1,6} \right)\) and \(\left( {1,2} \right)\) and focus at \(\left( {1,3} \right)\).
Solution:
Let’s graph the points we have, and go from there.
Ellipse  Math/Notes 
Domain: \(\left[ {1\sqrt{{15}},\,\,1+\sqrt{{15}}} \right]\) Range: \(\left[ {6,\,\,2} \right]\) 
Let’s first graph the points we have, and we can see that the ellipse is (barely!) vertical.We know that the endpoints \((–1,–6)\) and \((–1,2\)) are in fact vertices, and not covertices (since the focus is on the same line). The center is halfway between the two vertices, so it is \((1,2)\).
\(a=4\), so \({{a}^{2}}=1\). \(c=1\) (distance from center to a focus), so \({{c}^{2}}=1\). Since \({{c}^{2}}={{a}^{2}}{{b}^{2}}\), \({{b}^{2}}={{a}^{2}}{{c}^{2}}=161=15\). Thus, \(b=\sqrt{{15}}\), which we’ll need for the domain. Note that the covertices would be \(\left( {1\sqrt{{15}},2} \right)\) and \(\left( {1+\sqrt{{15}},2} \right)\).
The equation of the ellipse is \(\displaystyle \frac{{{{{\left( {x+1} \right)}}^{2}}}}{{15}}+\frac{{{{{\left( {y+2} \right)}}^{2}}}}{{16}}=\,\,1\).

Applications of Ellipses
The foci of ellipses are very useful in science for their reflective properties (sound waves, light rays and shockwaves, as examples), and are even used in medical applications. In fact, Kepler’s first law of planetary motion states that the path of a planet’s orbit models an ellipse with the sun at one focus, so the orbits of asteroids and other bodies are another elliptical application.
Problem:
Two girls are standing in a whispering gallery that is shaped like semielliptical arch. The height of the arch is 30 feet, and the width is 100 feet. How far from the center of the room should whispering dishes be placed so that the girls can whisper to each other? (Whispering dishes are places at the foci of an ellipse).
Solution:
Ellipse  Math/Notes 
We’ll put the center of the arch at \((0,0)\).
Since the width of the arch is 100 ft, \(a=50\). The height of the arch is 30 feet, so \(b=30\) (the height of the arch is only half of the minor axis of the ellipse). Therefore, we know the equation of the ellipse is \(\displaystyle \frac{{{{\text{x}}^{\text{2}}}}}{{\text{2500}}}\text{+}\frac{{{{\text{y}}^{\text{2}}}}}{{\text{900}}}\text{=}\,\,\text{1}\).
We need to get \(c\), so we can find the foci, since the whispering dishes are at the foci. Since \({{c}^{2}}={{a}^{2}}{{b}^{2}}\), \({{c}^{2}}={{50}^{2}}{{30}^{2}}=1600\), so \(c=40\).
Each girl would stand 40 feet from the center of the room.

Problem:
An ice rink is in the shape of an ellipse, and is 150 feet long and 75 feet wide. What is the width of the rink 15 feet from a vertex?
Ellipse  Math/Notes 
Let’s first find the equation of the ellipse, with the center at \((0,0)\). Since the major axis is 150, and the minor axis is 75, we have \(a=75\) and \(b=37.5\). From this, we know the equation of the ellipse is \(\displaystyle \frac{{{{x}^{2}}}}{{{{{75}}^{2}}}}+\frac{{{{y}^{2}}}}{{{{{37.5}}^{2}}}}=1\), or \(\displaystyle \frac{{{{x}^{2}}}}{{5625}}+\frac{{{{y}^{2}}}}{{1406.25}}=1\).
Now that we have the equation, we can plug in any \(x\) value to get the \(y\) value(s) on the ellipse; since we want the width of the ellipse 15 feet from the vertex, our \(x\) value is \(7515=60\). Plugging in 60 for \(x\), we get: \(\displaystyle \frac{{{{{60}}^{2}}}}{{5625}}+\frac{{{{y}^{2}}}}{{1406.25}}=1\); solving for \(y\), we get \(\displaystyle \pm \sqrt{{\left( {1\frac{{{{{60}}^{2}}}}{{5625}}} \right)\times 1406.25}}=\pm \,\,22.5\) (take positive only). Note that we need to take double 22.5 to get the whole width: the width of the rink 15 feet from a vertex is 45 feet.

Hyperbolas
A hyperbola sort of looks like two parabolas that point at each other, and is the set of points whose distances from two fixed points (the foci) inside the hyperbola is always the same, \({{d}_{1}}{{d}_{2}}=2a\). The distance \(2a\) is called the focal radii distance, focal constant, or constant difference, and \(a\) is the distance between the center of the hyperbola to a vertex.
The equation of a “horizontal” hyperbola (as shown below) that is centered on the origin \(\left( {0,0} \right)\) is \(\displaystyle \frac{{{x}^{2}}}{{{a}^{2}}}\frac{{{y}^{2}}}{{{b}^{2}}}=1\). The equation of a transformed horizontal hyperbola with center \((h,k)\) is \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\).
The length of the axis in which the hyperbola lies (called the transverse axis) is \(2a\), and this is along the \(x\)axis for a horizontal hyperbola. Again, the distance from the center of the hyperbola to a vertex is \(a\), so the vertices are at \(\left( \pm a,\,0 \right)\).
The length of the conjugate axis is \(2b\), and note that \(a\) does not have to be bigger than \(b\), like it does for an ellipse. (The distance from the center of the hyperbola to a covertex is \(b\)). Also note where the \(b\) is not on the hyperbola; it is on what we call the central rectangle (or fundamental rectangle) of the hyperbola (whose diagonals are asymptotes for the hyperbola). Thus, the conjugate axis is along the \(y\)axis for a horizontal hyperbola, and the covertices are at \(\left( 0,\,\pm b \right)\).
The asymptotes for a horizontal hyperbola centered at \(\left( {0,0} \right)\) are \(\displaystyle yk=\pm \frac{b}{a}\left( {xh} \right)\) ( \(\displaystyle \pm \frac{b}{a}\) are the slopes, or the square root of what’s under the \(y\) over the square root of what’s under the \(x\)). The asymptotes are the diagonals of the central rectangle of the hyperbola.
The focuses or foci always lie inside the curves on the major axis, and the distance from the center to a focus is \(c\). Thus, the foci are at \(\left( \pm c,\,\,0 \right)\) for a horizontal hyperbola (like an ellipse!), and it turns out that \({{a}^{2}}+{{b}^{2}}={{c}^{2}}\). (I like to remember that you always use the different sign for this equation: since ellipses have a plus sign in the equation \(\displaystyle \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\), they have a minus sign in \({{a}^{2}}{{b}^{2}}={{c}^{2}}\) ; since hyperbolas have a minus sign in the equation \(\displaystyle \frac{{{x}^{2}}}{{{a}^{2}}}\frac{{{y}^{2}}}{{{b}^{2}}}=1\), they have a plus sign in \({{a}^{2}}+{{b}^{2}}={{c}^{2}}\).)
Sometimes you will be asked to get the eccentricity of a hyperbola \(\displaystyle \frac{c}{a}\), which is a measure of how “straight” or “stretched” the hyperbola is.
Note also that, like for an ellipse, the focal width (focal chord, or focal rectum) of an ellipse is \(\displaystyle \frac{{2{{b}^{2}}}}{a}\); this the distance perpendicular to the major axis that goes through the focus.
Here are the two different “directions” of hyperbolas and the generalized equations for each:
Horizontal Hyperbola
(\({{x}^{2}}\) comes first) 
Vertical Hyperbola
(\({{y}^{2}}\) comes first) 
At \(\displaystyle \left( {0,0} \right):\,\,\,\,\,\frac{{{{x}^{2}}}}{{{{a}^{2}}}}\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1\) General: \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\) \({{a}^{2}}+{{b}^{2}}={{c}^{2}}\) Center: \(\left( {h,\,k} \right)\) Foci: \(\left( {h\pm c,\,\,k} \right)\) Vertices: \(\left( {h\pm a,\,\,k} \right)\) CoVertices: \(\left( {h,\,\,k\pm b} \right)\) Length of Transverse Axis: \(2a\) Length of Conjugate Axis: \(2b\) Asymptotes: \(\displaystyle yk=\pm \frac{b}{a}\left( {xh} \right)\) 
At \(\displaystyle \left( {0,0} \right):\,\,\,\,\,\frac{{{{y}^{2}}}}{{{{a}^{2}}}}\frac{{{{x}^{2}}}}{{{{b}^{2}}}}=1\): General: \(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\) \({{a}^{2}}+{{b}^{2}}={{c}^{2}}\) Center: \(\left( {h,\,\,k} \right)\) Foci: \(\left( {h,\,\,k\pm c} \right)\) Vertices: \(\left( {h,\,\,k\pm a} \right)\) CoVertices: \(\left( {h\pm b,\,\,k} \right)\) Length of Transverse Axis: \(2a\) Length of Conjugate Axis: \(2b\) Asymptotes: \(\displaystyle yk=\pm \frac{a}{b}\left( {xh} \right)\)

You also may have to complete the square to be able to graph an hyperbola, like we did here for a circle. (And since you always have to have a “1” after the equal sign, you may have to divide all terms by the constant on the right, if it isn’t “1”). Remember, for the conic to be a hyperbola, the coefficients of the \({{x}^{2}}\) and \({{y}^{2}}\) must have different signs.
Let’s put it all together and graph some hyperbolas:
Problem:
Identify the center, vertices, foci, and equations of the asymptotes for the following hyperbolas; then graph: (a) \(9{{x}^{2}}16{{y}^{2}}144=0\) (b) \(\displaystyle \frac{{{\left( y+3 \right)}^{2}}}{4}\frac{{{\left( x2 \right)}^{2}}}{36}=1\).
Solution:
It’s typically easier to graph the hyperbola first, and then answer the questions.
Hyperbola  Math/Notes 
(a) \(9{{x}^{2}}16{{y}^{2}}144=0\), which is \(\displaystyle \frac{{{{x}^{2}}}}{{16}}\frac{{{{y}^{2}}}}{9}=1\)
Domain: \(\left( {\infty ,4} \right]\cup \left[ {4,\infty } \right)\) Range: \(\left( {\infty ,\infty } \right)\)
Notice that the vertices and foci lie are along the horizontal line \(y=0\), and the length of the transverse axis is \(2a=8\). The length of the conjugate axis is \(2b=6\). 
We first need to get our equation into the form of hyperbola by adding 144 to both sides, and then dividing all terms by 144: \(\displaystyle \frac{{9{{x}^{2}}}}{{144}}\frac{{16{{y}^{2}}}}{{144}}=\frac{{144}}{{144}}\), or \(\displaystyle \frac{{{{x}^{2}}}}{{16}}\frac{{{{y}^{2}}}}{9}=1\). We will use the equation \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since the \(x\) comes first (horizontal).
This would make \({{a}^{2}}=16\), so \(a=4\). Since the hyperbola’s center is \((0,0)\), the vertices are \((4,0)\) and \((4,0)\). \({{b}^{2}}=9\), so \(b=3\). Thus, the covertices are \((0,3)\) and \((0,3)\). Now we can construct our central rectangle; we use \(a\) and \(b\) to create it.
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}+{{b}^{2}}=9+16=25\). Thus, \(c=5\) and the foci are \(\displaystyle \left( {\pm \,5,0} \right)\).
The equation of the asymptotes (which go through the corners of the central rectangle) are \(\displaystyle yk=\pm \frac{{b\text{ (rise)}}}{{a\text{ (run)}}}\left( {xh} \right)\), or \(\displaystyle y=\pm \frac{3}{4}x\). (Remember to use the square root of what’s under the \(y\) for the numerator of the slope, and the square of what’s under the \(x\) for the denominator.) 
(b) \(\displaystyle \frac{{{{{\left( {y+3} \right)}}^{2}}}}{4}\frac{{{{{\left( {x2} \right)}}^{2}}}}{{36}}=1\)
Domain: \(\left( {\infty ,\infty } \right)\) Range: \(\left( {\infty ,5} \right]\cup \left[ {1,\infty } \right)\)
Notice that the vertices and foci are along the vertical line \(x=2\), and the length of the transverse axis is \(2a=4\). The length of the conjugate axis is \(2b=12\). 
We will use equation: \(\displaystyle \frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since the \(y\) comes first (vertical). We see that the center of the hyperbola is at \((2,–3)\), so we can first plot that point.
\({{a}^{2}}=4\), so \(a=2\). Since the center is \((2,–3)\), the vertices are \((2,–3–2)\) and \((2,–3+2)\), or \((2,5)\) and \((2,1)\). \({{b}^{2}}=36\), so \(b=6\). Thus, the covertices are \((26,3)\) and \((2+6,3)\) or \((4,3)\) and \((8,3)\).
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}+\,\,{{b}^{2}}=4+36=40\). Thus, \(c=\sqrt{{40}}\text{ (or }\,2\sqrt{{10}})\), and the foci are \(\displaystyle \left( {2,3\,\pm \sqrt{{40}}} \right)\).
The equation of the asymptotes (which go through the corners of the central rectangle) are \(\displaystyle yk=\pm \frac{{a\text{ (rise)}}}{{b\text{ (run)}}}\left( {xh} \right)\), or \(\displaystyle y+3=\pm \frac{2}{6}\left( {x2} \right)\) or \(\displaystyle y+3=\pm \frac{1}{3}\left( {x2} \right)\). (Remember to use the square root of what’s under the \(y\) for the numerator of the slope, and the square of what’s under the \(x\) for the denominator.)

Here’s one where you have to Complete the Square to be able to graph the hyperbola:
Problem:
Identify the center, vertices, foci, and equations of the asymptotes for the following hyperbola; then graph: \(49{{y}^{2}}25{{x}^{2}}+98y100x+1174=0\).
Solution:
Hyperbola  Math/Notes 
\(49{{y}^{2}}25{{x}^{2}}+98y100x+1174=0\)
\(\displaystyle \begin{array}{c}49{{y}^{2}}+98y25{{x}^{2}}100x=1174\\49\left( {{{y}^{2}}+2y} \right)25\left( {{{x}^{2}}+4x} \right)=1174\\49\left( {{{y}^{2}}+2y+\underline{{\,\,\,\,\,\,}}\,} \right)25\left( {{{x}^{2}}+4x+\underline{{\,\,\,\,\,\,}}\,} \right)=1174+\underline{{\,\,\,\,\,\,}}\\49\left( {{{y}^{2}}+2y+\color{#2E8B57}{{\underline{{{{{\left( 1 \right)}}^{2}}}}}}\,} \right)25\left( {{{x}^{2}}+4x+\color{#2E8B57}{{\underline{{{{{\left( 2 \right)}}^{2}}}}}}\,} \right)=\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1174+\color{#2E8B57}{{\underline{{49{{{\left( 1 \right)}}^{2}}25{{{\left( 2 \right)}}^{2}}}}}}\\49{{\left( {y+1} \right)}^{2}}25{{\left( {x+2} \right)}^{2}}=1225\end{array}\) \(\displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{{49}}\frac{{{{{\left( {y+1} \right)}}^{2}}}}{{25}}=1\)
Domain: \(\left( {\infty ,9} \right]\cup \left[ {5,\infty } \right)\) Range: \(\left( {\infty ,\infty } \right)\) 
We need to first complete the square so we can get the equation in hyperbola form. Put all the \(x\)’s and \(y\)’s together with the constant term on the other side. Watch the negative signs; it turns to be a horizontal hyperbola, with the coefficient of the \({{x}^{2}}\) being positive (we ended up dividing all terms by –1225).
We will use equation \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\), since \(x\) comes first (horizontal hyperbola). We see that the center of the hyperbola is at \((–2,–1)\), so we can first plot that point.
\({{a}^{2}}=49\), so \(a=7\). Since the center is \((–2,–1)\), the vertices are \((–27,–1)\) and \((–2+7,–1)\), or \((–9,–1)\) and \((5,–1)\). \({{b}^{2}}=25\), so \(b=5\). The covertices are \((–2,15)\) and \((–2,1+5)\) or \((–2,–6)\) and \((–2,4)\).
Now let’s find the foci: \({{c}^{2}}={{a}^{2}}+{{b}^{2}}=49+25=74\). \(c=\sqrt{{74}}\) , and the foci are \(\displaystyle \left( {2\pm \sqrt{{74}},\,\,1\,} \right)\).
The equation of the asymptotes (which go through the corners of the central rectangle) are \(\displaystyle yk=\pm \frac{{b\text{ (rise)}}}{{a\text{ (run)}}}\left( {xh} \right)\), or \(\displaystyle y+1=\pm \frac{5}{7}\left( {x+2} \right)\).

Writing Equations of Hyperbolas
You may be asked to write an equation from either a graph or a description of a hyperbola:
Hyperbola  Math/Notes 
Write the equation of the hyperbola:

We can see that the center of the hyperbola is \(\left( {2,5} \right)\), the transverse axis length (\(2a\)) is 6, and the conjugate axis length (\(2b\)) is also 6. Thus, \(a=3\), and \(b=3\).
Since the hyperbola is horizontal, we’ll use the equation \(\displaystyle \frac{{{{{\left( {xh} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {yk} \right)}}^{2}}}}{{{{b}^{2}}}}=1\). Plug in our values, and we get \(\displaystyle \frac{{{{{\left( {x2} \right)}}^{2}}}}{9}\frac{{{{{\left( {y+5} \right)}}^{2}}}}{9}=1\). 
Problem:
Find the equation of the hyperbola where the difference of the focal radii is 6, and the endpoints of the conjugate axis are \(\left( {2,8} \right)\) and \(\left( {2,2} \right)\).
Solution:
We probably don’t even need to graph this hyperbola, since we’re basically given what \(a\) and \(b\) are. Remember that the difference of the focal radii is \(2a\), so \(a=3\).
Since the endpoints of the conjugate axis are along a vertical line, we know that the hyperbola is horizontal, and the covertices are \(\left( {2,8} \right)\) and \(\left( {2,2} \right)\). From this information, we can get the center (midpoint between the covertices), which is \(\left( {2,3} \right)\) and the length of the minor axis (\(2b\)), which is 10. So \(b=5\). (Draw the points first if it’s difficult to see).
The equation of the ellipse then is \(\displaystyle \frac{{{\left( x+2 \right)}^{2}}}{9}\frac{{{\left( y3 \right)}^{2}}}{25}=\,\,1\).
Problem:
Find the equation of the hyperbola where one of the vertices is at \(\left( {3,2} \right)\), and the asymptotes are \(\displaystyle y2=\pm \frac{2}{3}\left( {x3} \right)\).
Solution:
Let’s try to graph this one, since it’s hard to tell what we know about it!
Hyperbola  Math/Notes 
We can see from the equation of the asymptotes that the center of the hyperbola is \(\left( {3,2} \right)\). Graph this center and also graph the vertex that is given to see that the hyperbola is horizontal.
We can see that \(a\) (difference between center and vertex) is 6. So far then we have: \(\displaystyle \frac{{{{{\left( {x3} \right)}}^{2}}}}{{{{6}^{2}}}}\frac{{{{{\left( {y2} \right)}}^{2}}}}{{{{b}^{2}}}}=\,\,1\) We also see from the asymptotes equation that their slope is \(\displaystyle \pm \frac{2}{3}\). We actually don’t even need to draw them, since we know in our case, since it’s a horizontal hyperbola, we’ll have \(\displaystyle y2=\pm \frac{{b\text{ (rise)}}}{{6\text{ (run)}}}\left( {x3} \right)\) (rise is the square root of what’s under the \(y\); run is the square root of what’s under the \(x\)) from the equation of the hyperbola above. Now we can set up a proportion for the asymptote slopes: \(\displaystyle \frac{b}{6}=\frac{2}{3}\); by cross multiplying, we get \(b=4\). The equation of the hyperbola is: \(\displaystyle \frac{{{{{\left( {x3} \right)}}^{2}}}}{{{{6}^{2}}}}\frac{{{{{\left( {y2} \right)}}^{2}}}}{{{{4}^{2}}}}=\,\,1\), or \(\displaystyle \frac{{{{{\left( {x3} \right)}}^{2}}}}{{36}}\frac{{{{{\left( {y2} \right)}}^{2}}}}{{16}}=\,\,1\). 
Applications of Hyperbolas
Like ellipses, the foci of hyperbolas are very useful in science for their reflective properties, and hyperbolic properties are often used in telescopes. They are also used to model paths of moving objects, such as alpha particles passing the nuclei of atoms, or a spacecraft moving past the moon to the planet Venus.
Problem:
A comet’s path (as it approaches the sun) can be modeled by one branch of the hyperbola \(\displaystyle \frac{{{y}^{2}}}{1096}\frac{{{x}^{2}}}{41334}=\,\,1\), where the sun is at the focus of that part of the hyperbola. Each unit of the coordinate system is 1 million miles. (a) Find the coordinates of the sun (assuming it is at the focus with nonnegative coordinates). Round to the nearest hundredth. (b) How close does the comet come to the sun?
Solution:
Again, it’s typically easier to graph the hyperbola first, and then answer the questions.
Hyperbola  Math/Notes 
We’ll put the center of the hyperbola at \((0,0)\), and only work with the positive branch. The hyperbola is vertical since the \({{y}^{2}}\) comes before the \({{x}^{2}}\). \({{a}^{2}}=1096\) and \({{b}^{2}}=41334\).
(a) Since the sun is at a focus, we can use the equation \({{c}^{2}}={{a}^{2}}+{{b}^{2}}\) and take the positive value of \(c\), which is \(\sqrt{{1096+41334}}\approx 205.99\). The coordinates of the sun is \(\left( {0,205.99} \right)\), where each unit is in millions of miles.
(b) The closest the comet gets the sun as when the comet is at the vertex, which is \(\left( {0,a} \right)\), or \(\left( {0,33.11} \right)\). The closest the comet gets to the sun is about \(20633=173\) million miles.

Problem:
Two buildings in a shopping complex are shaped like a branches of the hyperbola \(729{{x}^{2}}1024{{y}^{2}}746496=0\), where \(x\) and \(y\) are in feet. How far apart are the buildings at their closest part?
Solution:
Let’s try this one without drawing it, since we know that the closest points of a hyperbola are where the vertices are, and the buildings would be \(2a\) feet apart.
By doing a little algebra (adding 746496 to both sides and then dividing all terms by 746496) we see that the equation in hyperbolic form is \(\displaystyle \frac{{{x}^{2}}}{1024}\frac{{{y}^{2}}}{729}=1\). So \(a=\sqrt{{1024}}=32\). The building are 32 x 2 = 64 feet apart at their closest part.
Problem:
Two radar sites are tracking an airplane that is flying on a hyperbolic path. The first radar site is located at \(\left( {0,0} \right)\), and shows the airplane to be 200 meters away at a certain time. The second radar site, located 160 miles east of the first, shows the airplane to be 100 meters away at this same time. Find the coordinates of all possible points where the airplane could be located. (Find the equation of the hyperbola where the plane could be located).
Solution:
Let’s draw a picture first and remember that the constant difference for a hyperbola is always \(2a\). The plane’s path is actually on one branch of the hyperbola; let’s create a horizontal hyperbola, so we’ll use the equation \(\displaystyle \frac{{{\left( xh \right)}^{2}}}{{{a}^{2}}}\frac{{{\left( yk \right)}^{2}}}{{{b}^{2}}}=1\):
Hyperbola  Math/Notes 
We know that the distance from the “leftmost” focus to the plane (hyperbola) is 200 meters, and the distance from the “rightmost” focus to the plane (hyperbola) is 100 meters.
This is actually the “constant difference” of the hyperbola, which is \(2a\). \(200100=2a\), or \(a=50\). Thus, \({{a}^{2}}=2500\). We also know that \(2c\) (distance between foci) \(=160\), so \(c=80\).
Since \({{c}^{2}}={{a}^{2}}+{{b}^{2}}\), we can obtain \({{b}^{2}}:\,{{b}^{2}}={{c}^{2}}{{a}^{2}}={{80}^{2}}{{50}^{2}}=3900\).
In the model, the center of the hyperbola is at \((80,0)\), so the path of the airplane follows the hyperbola\(\displaystyle \frac{{{{{\left( {x80} \right)}}^{2}}}}{{2500}}\frac{{{{y}^{2}}}}{{3900}}=1\) .

Problem:
Alpha particles are deflected along hyperbolic paths when they are directed towards the nuclei of gold atoms. If an alpha particle gets as close as 10 units to the nucleus along a hyperbolic path with asymptote \(\displaystyle y=\frac{2}{5}x\), what is the equation of its path?
Solution:
Let’s draw a picture first and make the nucleus the center of the hyperbola at \(\left( {0,0} \right)\).
Hyperbola  Math/Notes 
We can put the nucleus at \((0,0)\), and the asymptotes at \(\displaystyle y=\pm \frac{2}{5}x\).
From the drawing, we see that the closest the hyperbolic alpha particle gets to the nucleus is at \((0,a)\); thus \(a=10\).
Now we have to figure out what \(b\) is; we need to use the equation of the asymptote to do this. We do know that \(\displaystyle \frac{b}{a}=\frac{2}{5}\) (formula of asymptotes for horizontal hyperbola), and that \(a=10\). By cross multiplying with \(\displaystyle \frac{b}{{10}}=\frac{2}{5}\), we get \(b=4\).
Therefore the path of the alpha particle follows the hyperbola .\(\displaystyle \frac{{{{x}^{2}}}}{{100}}\frac{{{{y}^{2}}}}{{16}}=1\)

Identifying the Conic
Sometimes you are given an equation or a description of a conic, and asked to identify the conic. Remember these rules:
 \({{x}^{2}}\) with other \(y\)’s (and maybe \(x\)’s), or \({{y}^{2}}\) with other \(x\)’s (and maybe \(y\)’s): parabola
 \({{x}^{2}}\) and \({{y}^{2}}\) with same coefficients and + sign: circle
 \({{x}^{2}}\) and \({{y}^{2}}\) with same coefficients and – sign: hyperbola
 \({{x}^{2}}\) and \({{y}^{2}}\) with different coefficients and + sign: ellipse
 \({{x}^{2}}\) and \({{y}^{2}}\) with different coefficients and – sign: hyperbola
Here are some examples; I always find it’s easier to work/graph these on graph paper to see what’s going on:
Identify the Conic  Solution 
Identify these conics:
(a) \(16{{y}^{2}}=4{{x}^{2}}+64\)
(b) \(6{{x}^{2}}6{{y}^{2}}=54\)
(c) \(\displaystyle {{x}^{2}}+{{y}^{2}}=4xy+4\)
(d) \({{x}^{2}}2y=x+3\) 
(a) Get all variables on one side: \(4{{x}^{2}}+16{{y}^{2}}=64\). Since the coefficients of \({{x}^{2}}\) and \({{y}^{2}}\) are different, but we have a \(+\) sign, it’s an ellipse. (We would end up with \(\displaystyle \frac{{{{x}^{2}}}}{{16}}+\frac{{{{y}^{2}}}}{4}=1\).)
(b) The coefficients of \({{x}^{2}}\) and \({{y}^{2}}\) are the same, but we have a \(\) sign between them, it’s a hyperbola. (We would end up with \(\displaystyle \frac{{{{x}^{2}}}}{9}\frac{{{{y}^{2}}}}{9}=1\).)
(c) Since the coefficients of \({{x}^{2}}\) and \({{y}^{2}}\) are the same, we have a circle. We’d have to complete the square to get it in \({{\left( {xh} \right)}^{2}}+{{\left( {yk} \right)}^{2}}={{r}^{2}}\) form.
(d) Since we have \({{x}^{2}}\) with other \({{y}^{2}}\)’s and \({{x}^{2}}\)’s, we have a parabola. We’d have to complete the square to get it into \(y=a{{\left( {xh} \right)}^{2}}+k\) (horizontal) form.

For the following, write the equation of the conic, using the given information:
Conic  Solution 
Ellipse with foci \(\left( {0,2\sqrt{{13}}} \right),\left( {0,2\sqrt{{13}}} \right)\), and focal constant 26  The focal constant is the same as the constant sum, and is defined as \(2a\) for an ellipse (so \(a=13\)). Since the foci are “up and down”, we know it’s a vertical ellipse, so we have \(\displaystyle \frac{{{{x}^{2}}}}{{{{b}^{2}}}}+\frac{{{{y}^{2}}}}{{{{{13}}^{2}}}}=1\).
To get \(b\), we can use \({{c}^{2}}={{a}^{2}}{{b}^{2}}\), and we know that \({{c}^{2}}={{\left( {2\sqrt{{13}}} \right)}^{2}}=52\). \({{b}^{2}}={{a}^{2}}{{c}^{2}}={{13}^{2}}52=117\). The equation of the ellipse is \(\displaystyle \frac{{{{x}^{2}}}}{{117}}+\frac{{{{y}^{2}}}}{{{{{13}}^{2}}}}=1\), or \(\displaystyle \frac{{{{x}^{2}}}}{{117}}+\frac{{{{y}^{2}}}}{{169}}=1\). 
The equations of the asymptotes are \(\displaystyle y=\pm 3\left( {x+6} \right)2\), and the length of the horizontal conjugate axis is 10  The length of the horizontal conjugate axis is \(2b\) for a hyperbola, and for this problem, \(2b=10\), so \(b=5\). Since the conjugate axis is horizontal, we know we have a vertical hyperbola.
We know the center of the hyperbola is \((6,2)\) from the asymptotes. Thus, we have \(\displaystyle \frac{{{{{\left( {y+2} \right)}}^{2}}}}{{{{a}^{2}}}}\frac{{{{{\left( {x+6} \right)}}^{2}}}}{{{{5}^{2}}}}=1\). To get \(a\), we can use the equation of the asymptotes: we know that \(\displaystyle \frac{a}{b}\)(what’s under the \(y\) over what’s under the \(x\)) \(=3\). \(\displaystyle \frac{a}{5}=\frac{3}{1}\), or \(a=15\). The equation of the hyperbola is \(\displaystyle \frac{{{{{\left( {y+2} \right)}}^{2}}}}{{{{{15}}^{2}}}}\frac{{{{{\left( {x+6} \right)}}^{2}}}}{{{{5}^{2}}}}=1\). 
The set of points that are equidistant from a fixed point \((–3,5)\) as they are from a fixed line \(x=8\).  By definition, we know this is a parabola, and the focus is at \((3,5)\). Since the vertex is halfway between the focus and the directrix (\(x=8\)), we know the vertex is at \(\displaystyle \left( {\frac{{3+8}}{2},5} \right),\text{ or }\left( {2.5,5} \right),\) and the parabola is horizontal and opens up to the left (draw it!).
We also know that \(p\) (distance from vertex to focus) is \(82.5=5.5\), so the equation is \(\displaystyle x=\frac{1}{{4p}}{{\left( {yk} \right)}^{2}}+h\), or \(\displaystyle x=\frac{1}{{22}}{{\left( {y5} \right)}^{2}}+2.5\). 
Foci at \(\left( {2,4\pm \sqrt{5}} \right)\) and endpoints of an axis are at \(\left( {2\pm \sqrt{3},4} \right)\)  It’s best to graph this one! We see that it must be a vertical ellipse, since we’re talking about foci and endpoints, and the foci are vertical. We can see that \(b=\sqrt{3}\) since that’s what’s added and subtracted horizontally to the center of the ellipse \((2,4)\) (draw it!).
We have \(\displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{3}+\frac{{{{{\left( {y4} \right)}}^{2}}}}{{{{a}^{2}}}}=1\). To get \(a\), we can use \({{c}^{2}}={{a}^{2}}{{b}^{2}}\), and we know that \({{c}^{2}}\) is \({{\left( {\sqrt{5}} \right)}^{2}}=5\). So, \({{a}^{2}}={{c}^{2}}+{{b}^{2}}=5+3=8\). The equation of the ellipse is \(\displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{3}+\frac{{{{{\left( {y4} \right)}}^{2}}}}{8}=1\). 
Learn these rules, and practice, practice, practice!
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On to Systems of NonLinear Equations – you are ready!