# Conics: Circles, Parabolas, Ellipses, and Hyperbolas

This section covers:

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

Note that we can also write equations for circles, ellipses, and hyperbolas in terms of cos and sin, and other trigonometric functions; 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:    .

If we were to solve for y in terms of x (for example, to put in the graphing calculator), we’d get:      .

Here are graphs of sample circles, with their domains and ranges:

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

Solution:

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:

## 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  $${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=61$$  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 (–2, –8),  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:

Here’s another type of problem you might see:

Problem:

The lines  $$y=\frac{4}{3}x-\frac{5}{3}$$   and   $$y=-\frac{4}{3}x-\frac{{13}}{3}$$   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:

## Applications of Circles

Problem:

A pizza delivery area can be represented by a circle, and extends to the points (0, 18) and (–6, 8) (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:

# 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).

We remember that a parabola is in the form  $$y=a{{\left( x-h \right)}^{2}}+k$$, where (h, k) 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  $$y-k=a{{\left( {x-h} \right)}^{2}}$$  or  $$b\left( {y-k} \right)={{\left( {x-h} \right)}^{2}}+k$$, where  $$b=\frac{1}{a}$$.

Parabolas can also be in the form  $$x=a{{\left( y-k \right)}^{2}}+h$$, where (h, k) is the vertex, and “y = k”  is the LOS; this is a “horizontal” parabolaThe line of symmetry (LOS) is a line that divides the parabola into two parts that are mirror images of each other.

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).

So 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.  You can use this information to help you draw the parabola, if you know the distance from the vertex to the focus (p).

Here is a parabola with center (0, 0):

IIf the vertex is at the origin (0, 0), the equation of the parabola is  $$y=a{{x}^{2}}$$, and  $$a=\frac{1}{4p}$$;  if you do the algebra, it follows that that  $$p=\frac{1}{4a}$$.

For example, if p = 4 (length of focus to vertex),  the equation of the parabola would be  $$y=\frac{1}{4\left( 4 \right)}{{x}^{2}}\,\,=\,\,y=\frac{1}{16}{{x}^{2}}$$.

Note that the above graph shows a parabola that has a positive value before the  $${{x}^{2}}$$.

Here are the four different “directions” of parabolas and the generalized equations for each:

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) $$y-4=2{{\left( x-3 \right)}^{2}}$$  (b)  $$x=-\frac{{{y}^{2}}}{2}+2y+4$$.

Solution:

It’s typically easier to graph the parabola first, and then answer the questions.

## Writing Equations of Parabolas

Problem:

Write the equation of a parabola with a vertex of ( –2, 4) and focus point (0, 4).   Also find the domain and range of the parabola.

Solution:

Here’s another parabola problem that’s a bit tricky:

Problem:

Write the equation and graph a parabola with focus ( –2, –7) that opens to the right, and contains the point (6, –1).

Solution:

## 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  $$\frac{1}{32}{{x}^{2}}$$  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 (0, 0).  Since we know that the equation of a parabola is  $$y=a{{x}^{2}}$$, where  $$a=\frac{1}{4p}$$  and  $$p=\frac{1}{4a}$$, then for  $$\frac{1}{32}{{x}^{2}}$$, we have  $$\frac{1}{32}=\frac{1}{4p}$$.  We can either cross multiply or just do mental math to see that p = 8.  So 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 (0, 0).

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 (0, 0).

# 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 co-vertices).

The equation of a “horizontal” ellipse that is centered on the origin (0, 0) is  $$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$$  (what’s under the x is larger than what’s under the y).  The 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 co-vertex is b,  so the co-vertices 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.  So 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  $$\frac{c}{a}$$, 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  $$\frac{{2{{b}^{2}}}}{a}$$;  this the distance perpendicular to the major axis that goes through the focus.

Here is a horizontal ellipse; we will also look at vertical and transformed ellipses below.

Here are the two different “directions” of ellipses and the generalized equations for each:

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, co-vertices, foci, and domain and range for the following ellipses; then graph: (a)  $$9{{x}^{2}}+49{{y}^{2}}=441$$    (b)   $$\frac{{{\left( x+3 \right)}^{2}}}{4}+\frac{{{\left( y-2 \right)}^{2}}}{36}=1$$.

Solution:

It’s typically easier to graph the ellipse first, and then answer the questions:

Here’s one where you have to Complete the Square to be able to graph the ellipse:

Problem:

Identify the vertices, co-vertices, foci and domain and range for the following ellipse; then graph $$4{{x}^{2}}+{{y}^{2}}+24x+2y=-33$$.

Solution:

## 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 (h, k)  is at (4, –3).

Since the ellipse is horizontal, we’ll use the equation  $$\frac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$$.  Plug in our values for a, b, h, and k, and we get  $$\frac{{{\left( x-4 \right)}^{2}}}{25}+\frac{{{\left( y+3 \right)}^{2}}}{4}=1$$.  Not too bad!

Problem:

Find the equation of this ellipse, graph, and find the domain and range:  Endpoints of major or minor axis at (–1, –6) and (–1, 2) and focus at (–1, –3).

Solution:

Let’s graph the points we have, and go from there.

## 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 semi-elliptical 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:

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?

Solution:

## 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 (0, 0) is  $$\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$$.   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 co-vertex 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).   So the conjugate axis is along the y-axis for a horizontal hyperbola, and the co-vertices are at  $$\left( 0,\,\pm b \right)$$.

The asymptotes for a horizontal hyperbola centered at (0, 0) are  $$y=\pm \frac{b}{a}x$$ ( $$\pm \frac{b}{a}$$  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.  So 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  $$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$$, they have a minus sign in  $${{a}^{2}}-{{b}^{2}}={{c}^{2}}$$ ; since hyperbolas have a minus sign in the equation  $$\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$$, they have a plus sign in  $${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$.)

Sometimes you will be asked to get the eccentricity of an hyperbola  $$\frac{c}{a}$$, 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  $$\frac{{2{{b}^{2}}}}{a}$$;  this the distance perpendicular to the major axis that goes through the focus.

Here is a horizontal hyperbola; we will also look at vertical and transformed hyperbolas below.

Here are the two different “directions” of hyperbolas and the generalized equations for each:

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)  $$\frac{{{\left( y+3 \right)}^{2}}}{4}-\frac{{{\left( x-2 \right)}^{2}}}{36}=1$$.

Solution:

It’s typically easier to graph the hyperbola first, and then answer the questions.

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:  $$49{{y}^{2}}-25{{x}^{2}}+98y-100x+1174=0$$.

Solution:

## Writing Equations of Hyperbolas

You may be asked to write an equation from either a graph or a description of a hyperbola:

Problem:

Write the equation of the hyperbola:

Solution:

We can see that the center of the hyperbola is (2, –5), the transverse axis length (2a)  is 6, and the conjugate axis length (2b) is also 6.   So a = 3, and b = 3.

Since the hyperbola is horizontal, we’ll use the equation  $$\frac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$$.   Plug in our values, and we get  $$\frac{{{\left( x-2 \right)}^{2}}}{9}-\frac{{{\left( y+5 \right)}^{2}}}{9}=1$$.

Problem:

Find the equation of the hyperbola where the difference of the focal radii is 6, and the endpoints of the conjugate axis are (–2, 8) and (–2, –2).

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 co-vertices are (–2, 8) and (–2, –2).  From this information, we can get the center (midpoint between the co-vertices), which is (–2, 3) and the length of the minor axis (2b), which is 10.  So b = 5.   (Draw the points first if it’s difficult to see).
So the equation of the ellipse is  $$\frac{{{\left( x+2 \right)}^{2}}}{9}-\frac{{{\left( y-3 \right)}^{2}}}{25}=\,\,1$$.

Problem:

Find the equation of the hyperbola where one of the vertices is at  $$\left( {-3,2} \right)$$, and the asymptotes are   $$y-2=\pm \frac{2}{3}\left( {x-3} \right)$$.

Solution:

Let’s try to graph this one, since it’s hard to tell what we know about it!

We can see from the equation of the asymptotes that the center of the hyperbola is  $$\left( {3,2} \right)$$.

Then we’ll graph this center and also graph the vertex that is given to see that the hyperbola is horizontal:

## 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  $$\frac{{{y}^{2}}}{1096}-\frac{{{x}^{2}}}{41334}=\,\,1$$,  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 non-negative 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.

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  $$\frac{{{x}^{2}}}{1024}-\frac{{{y}^{2}}}{729}=1$$.   So a  = $$\sqrt{1024}$$ = 32.  So 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 (0, 0), 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  $$\frac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$$:

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  $$y=\frac{2}{5}x$$, what is the equation of its path?

Solution:

Let’s draw a picture first and make the nucleus the center of the hyperbola at (0, 0).

# 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:

For the following, write the equation of the conic, using the given information:

Learn these rules, and practice, practice, practice!

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