Conics Part 1: Circles and Parabolas

Conics (circles, ellipses, parabolas, and hyperbolas) involves a set of curves that are formed by intersecting a plane and a double-napped 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. 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!

CONIC

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

HORIZONTAL

$ \displaystyle \begin{array}{c}{{\left( {x-h} \right)}^{2}}+{{\left( {y-k} \right)}^{2}}\\={{r}^{2}}\\\text{Point}\left( {h,k} \right)\text{is center of circle}\end{array}$

$ \displaystyle x=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}+h$

or

$ \displaystyle x-h=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}$

or

$ 4p\left( {x-h} \right)={{\left( {y-k} \right)}^{2}}$

Example has positive coefficient

$ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

$ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

Asymptotes:

$ \displaystyle y-k=\pm \frac{b}{a}\left( {x-h} \right)$

VERTICAL

No Change

$ \displaystyle y=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}+k$

or

$ \displaystyle y-k=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}$

 or

$ 4p\left( {y-k} \right)={{\left( {x-h} \right)}^{2}}$

Example has positive coefficient

$ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

$ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

Asymptotes:

$ \displaystyle y-k=\pm \frac{a}{b}\left( {x-h} \right)$

Other

Information

To get $ y$:

$ \begin{array}{c}y=\\\,\,\pm \sqrt{{{{r}^{2}}-{{{\left( {x-h} \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 a conic exists, if $ {{B}^{2}}-4AC<0$, it is a circle or ellipse, if $ {{B}^{2}}-4AC=0$, it is a parabola, and if $ {{B}^{2}}-4AC>0$, 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( {x-h} \right)}^{2}}+{{\left( {y-k} \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( {x-h} \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. Note that, after completing the square, we may not necessarily get a circle if the coefficients of $ {{x}^{2}}$ and $ {{y}^{2}}$ are not both positive 1, as we’ll see later.

Here are some examples:

Completing the Square Circle Problems Notes

Find the center and radius of the following circle:

$ {{x}^{2}}+{{y}^{2}}+8y-9=0$

 

$ \require {cancel} \displaystyle \begin{align}{{x}^{2}}+{{y}^{2}}+8y&=9\\{{x}^{2}}+\left( {{{y}^{2}}+8y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=9+\,\,\underline{{\,\,\,\,\,\,}}\\{{x}^{2}}+\left( {{{y}^{2}}+8y+\underline{{{{{\left( 4 \right)}}^{2}}\,}}} \right)&=9+\underline{{{{{\left( 4 \right)}}^{2}}}}\\{{x}^{2}}+{{\left( {y+\underline{4}} \right)}^{2}}&={{5}^{2}}\end{align}$

Move the constant to the right side, group the $ x$’s and $ y$’s’s together, and we are ready to complete the square.

 

Notice that we don’t have to complete the square for the $ x$ terms, since there’s no middle term. Divide the coefficient of the $ y$ term by 2 and square it to complete the square. Add the squared constant to the other side.

 

The equation is now in circle form. Note that $ {{x}^{2}}$ is the same as $ {{\left( {x-0} \right)}^{2}}$, so this equation is a circle with center $ (0,-4)$ and radius 5.

Find the center and radius of the following circle:

$ {{x}^{2}}+{{y}^{2}}-6x-12y-55=0$

 

$ \displaystyle \begin{align}{{x}^{2}}-6x+{{y}^{2}}-12y&=55\\\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( {x-3} \right)}^{2}}+{{\left( {y-6} \right)}^{2}}&=100\\{{\left( {x-3} \right)}^{2}}+{{\left( {y-6} \right)}^{2}}&={{10}^{2}}\end{align}$

Move the constant to the right side, group the $ x$’s and $ y$’s’s together, and we are ready to complete the square!

 

Divide the coefficients (both $ x$ and $ y$) of the middle terms by 2 and square them to complete the square. Add the squared constants to the other side.

 

The equation is now in circle form! This equation is a circle with center $ (3,6)$ and radius 10.

Complete the square and describe the graph:

$ {{x}^{2}}+{{y}^{2}}-4x+2y+5=0$

 

$ \displaystyle \begin{align}{{x}^{2}}-4x+{{y}^{2}}+2y&=-5\\\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( {x-2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}&=0\end{align}$

Move the constant to the right side to complete the square.

 

Divide the coefficients (both $ x$ and $ y$) of the middle terms by 2 and square them to complete the square. Add the squared constants to the other side.

 

The equation is in circle form, but our radius is 0!

 

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! Tricky!

Writing Equations of Circles

Sometimes you will have to come up with the equations of circle, or equations of tangents of circles.

Problem:

Write the equation of the line that is tangent to the circle $ {{\left( x-3 \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 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}$. Since the tangent line is perpendicular to this line, its slope is $ \displaystyle -\frac{5}{6}$.

Using this slope and point $ (–2,–8)$, we can use either the slope-intercept or point-slope method to get the equation; let’s use the slope-intercept:

$ \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:   

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( {-3-0} \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:

Circle Graph Solution
First plot 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( {18-13} \right)}}^{2}}}}=\sqrt{{34}}$.

The equation of circle is $ {{\left( {x+3} \right)}^{2}}+{{\left( {y-13} \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 with positive coefficients (“cup up”) and negative coefficients (“cup down”).

Vertical parabolas are in the form $ y=a{{\left( x-h \right)}^{2}}+k$, where $ \left( {h,\,k} \right)$ is the vertex and $ x=h$ is the axis of symmetry or line of symmetry (LOS). Note that this can also be written $ y-k=a{{\left( {x-h} \right)}^{2}}$ or $ b\left( {y-k} \right)={{\left( {x-h} \right)}^{2}}+k$, where $ \displaystyle b=\frac{1}{a}$. The line of symmetry (LOS) is a line that divides the parabola into two parts that are mirror images of each other. 

Parabolas can also be in the form $ x=a{{\left( y-k \right)}^{2}}+h$, where $ \left( {h,\,k} \right)$ is the vertex, and $ y=k$ is the LOS; this is a horizontal parabola. 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=$”), and for horizontal (sideways) parabolas, the directrix is a vertical line (“$ 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, focal chord or focal rectum; the length of this chord (focal width or focal diameter) is $ 4p$. This is cool! 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 vertical parabola with center $ \left( {0,\,0} \right)$:

If the vertex is at the origin $ \left( {0,\,0} \right)$, the equation of a vertical 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$”.

Note that in these formulas and subsequent examples, I assume $ p$ is a distance; thus, it is always positive (and so is $ \displaystyle a=\frac{1}{{4p}}$). Many teach using just one formula for positive and negative coefficients; thus, $ p$ is just the coefficient, and can be negative. Sorry for the confusion; either approach can be used.

Vertical Parabola Horizontal Parabola

Positive Coefficient

At $ \left( {0,0} \right):\,\,\,\,\,y=a{{x}^{2}}$

$ y=a{{\left( {x-h}\right)}^{2}}+k$  or  $ y-k=a{{\left( {x-h} \right)}^{2}}$

 $ \displaystyle y=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}+k$   or  $ \displaystyle y-k=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}$

or

$ 4p\left( {y-k} \right)={{\left( {x-h} \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( {y-k} \right)}^{2}}+h$  or  $ x-h=a{{\left( {y-k} \right)}^{2}}$

$ \displaystyle x=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}+h$    or    $ \displaystyle x-h=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}$

or

$ 4p\left( {x-h} \right)={{\left( {y-k} \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( {x-h} \right)}^{2}}+k$   or   $ y-k=-a{{\left( {x-h} \right)}^{2}}$

$ \displaystyle y=-\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}+k$    or    $ \displaystyle y-k=-\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}$

or

$ -4p\left( {y-k} \right)={{\left( {x-h} \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( {y-k} \right)}^{2}}+h$   or   $ x-h=-a{{\left( {y-k} \right)}^{2}}$

$ \displaystyle x=-\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}+h$    or    $ \displaystyle x-h=-\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}$

or

$ -4p\left( {x-h} \right)={{\left( {y-k} \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!

Parabola Problems and Graphs Solution

Identify the vertex, axis of symmetry, focus, equation of the directrix, and domain and range for the following parabola; then graph the parabola: $ \displaystyle y-4=\frac{1}{{16}}{{\left( {x-3} \right)}^{2}}$.

 

Domain:  $ \left( {-\infty ,\infty } \right)$    Range:  $ \left[ {4,\infty } \right)$

We see the equation is in the form $ \displaystyle y-k=\frac{1}{{4p}}{{\left( {x-h} \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$; for example, when $ \displaystyle x=11,\,y=\frac{1}{{16}}{{\left( {11-3} \right)}^{2}}+4=8$).

Identify the vertex, axis of symmetry, focus, equation of the directrix, and domain and range for the following parabola; then graph the parabola:  $ \displaystyle {{y}^{2}}-4y+2x-8=0$. (This is in standard or general form).

\begin{align}x-4&=-\frac{1}{2}\left( {{{y}^{2}}-4y} \right)\\x-4-\color{#117A65}
{{\underline{{\frac{1}{2}{{{\left( 2 \right)}}^{2}}}}}}&=\color{#117A65}{{-\frac{1}{2}}}\left( {{{y}^{2}}-4y+\color{#117A65}
{{\underline{{{{{\left( 2 \right)}}^{2}}}}}}} \right)\\x-6&=-.5{{\left( {y-2} \right)}^{2}}\end{align}

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 x-h=\left( – \right)\frac{1}{{4p}}{{\left( {y-k} \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 $ x-6=-.5{{\left( {y-2} \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$; for example, when $ \displaystyle x=4,\,y=\pm \sqrt{{\frac{{4-6}}{{-.5}}}}+2=0,4$).

Writing Equations of Parabolas

Here are some problems where we need to find equations of parabolas, given certain conditions.

Parabola Equation Problems

Solution

Write the equation of a parabola with a vertex of $ \left( {-2,4} \right)$ and focus point $ \left( {0,4} \right)$.

Graph:

Domain: $ \left[ {-2,\infty } \right)$     Range: $ \left( {-\infty ,\infty } \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 x-h=\frac{1}{{4p}}{{\left( {y-k} \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( {y-4} \right)}^{2}}$, or $ \displaystyle x+2=\frac{1}{8}{{\left( {y-4} \right)}^{2}}$. You can also write the parabola as $ \displaystyle x=\frac{1}{8}{{\left( {y-4} \right)}^{2}}-2$ or $ 8\left( {x+2} \right)={{\left( {y-4} \right)}^{2}}$.

Write the equation of the parabola with a focus of $ \left( {-2,4} \right)$ and a directrix of $ y=9$.

Graph:

Domain: $ \left( {-\infty ,\infty } \right)$     Range: $ \displaystyle \left( {-\infty ,6.5} \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 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 y-h=-\frac{1}{{4p}}{{\left( {x-k} \right)}^{2}}$, where $ p$ is focal length.

 

Since the length from the focus to the directrix is $ 5\,\,\,(9-4)$, 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 y-6.5=-\frac{1}{{4\left( {2.5} \right)}}{{\left( {x-\left( {-2} \right)} \right)}^{2}}$, or $ \displaystyle y-6.5=-\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( {y-6.5} \right)=-{{\left( {x+2} \right)}^{2}}$.

Here are more problems where we need to find an equation for a parabola; the second one is really tricky!

Parabola Graph Problems Solution

Write the equation of the parabola with a vertex of $ \left( {-3,-3} \right)$, contains the point $ \left( {1,-4} \right)$, and opens vertically.

Graph:

It’s best to first plot the points, so we can that the parabola opens down.

 

Since we know the vertex, we have $ \displaystyle y+3=-\frac{1}{{4p}}{{\left( {x+3} \right)}^{2}}$.

 

Put in the other point to get $ \displaystyle -4+3=-\frac{1}{{4p}}{{\left( {1+3} \right)}^{2}};\,\,p=4$:

 

The equation for the parabola is $ \displaystyle y+3=-\frac{1}{{16}}{{\left( {x+3} \right)}^{2}}$.

Write the equation of the parabola with a focus of $ \left( {-2,-7} \right)$ that opens to the right, parabola contains the point $ \left( {6,-1} \right)$.

Graph:

We know that the equation for a parabola that opens up to the right is $ \displaystyle x-h=\frac{1}{{4p}}{{\left( {y-k} \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( {-2-p,-7} \right)$. From this, we know that $ h=-2-p$ and $ k=-7$.

 

We can now plug in the point $ \left( {6,-1} \right)$ for $ x$ and $ y$, and $ \left( {-2-p,-7} \right)$ for $ \left( {h,k} \right)$, and solve for $ p$:

$ \require {cancel} \begin{array}{c}6-\left( {-2-p} \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}}+8p-9=0\,\,;\,\,\,\,\,\,\,\left( {p+9} \right)\left( {p-1} \right)=0\,\\p=\cancel{{-9}},\,1\end{array}$

 

Since $ p=1$, the vertex is at $ \left( {-2-1,-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 real-world 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; this is the focal width).

 

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 bulb in a searchlight is placed at the focus of a parabolic mirror, which is .75 feet from the vertex. This causes the rays of light from the bulb to bounce off of the mirror as parallel rays, providing a concentrated beam of light.

(a) What is the equation of the parabola, if the focal diameter of the bulb is 3 feet?  (b) Suppose, in another searchlight, the focal diameter is 4 feet. If the depth of both searchlights is 5 feet, how much greater is the width of the opening of this larger light than the searchlight with a focal diameter of 3 feet?

Solution:
Let’s graph this particular parabola, again putting the vertex at $ \left( {0,0} \right)$. As in other problems, it’s easier to make a vertical parabola.

Parabola Pictures Solution
(a)
(b)
(a)  The bulb is .75 feet from the vertex and, since the focal diameter is 3 feet, this point is 1.5 feet from either side of the parabola (the focal width). Thus, we know $ \left( {1.5,.75} \right)$ is a point on the parabola. We can use the form $ y=a{{x}^{2}}$, where $ \displaystyle a=\frac{1}{{4p}}$, to get its equation:

$ \begin{align}y&=a{{x}^{2}}\\.75&=a{{\left( {1.5} \right)}^{2}};\,\,\,a=\frac{1}{3}\\y&=\frac{1}{3}{{x}^{2}}\end{align}$

(We also could have used the fact that $ 4p=$  the focal diameter (3), so $ \displaystyle p=\frac{3}{4}$. Then, since $ \displaystyle y=\frac{1}{{4p}}{{x}^{2}}$, we have $ \displaystyle y=\frac{1}{3}{{x}^{2}}$.)

(b)  Since the focal diameter on the new light is 4 instead of 3, we know $ p=1$ (since $ 4p=4$). Thus, the equation of the parabola is: $ \displaystyle y=\frac{1}{{4\left( 1 \right)}}{{x}^{2}}=\frac{1}{4}{{x}^{2}}$.

Since the depth of both of these searchlights is 5, so we can plug in 5 for the $ y$-values to get the $ x$-values back (half of the widths at that point).

$ \displaystyle y=\frac{1}{3}{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{1}{4}{{x}^{2}}$

$ \displaystyle 5=\frac{1}{3}{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5=\frac{1}{4}{{x}^{2}}$

$ \displaystyle x\approx 3.873\,\,\,\,\,\,\,\,\,\,\,\,\,x\approx 4.472$

Double these values to get approximate total widths of 7.746 and 8.944, respectively. The difference in the widths is about 1.198 feet.

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.

On to Conics Part 2: Ellipses and Hyperbolas