Conics: Part 2: Ellipses and Hyperbolas

Note that Part 1 of Conics (Circles and Parabolas) is here.

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!

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 constant: ${{d}_{1}}+{{d}_{2}}=2a$. 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 ${{d}_{2}}$); thus, the constant sum is the distance between the vertices. Can you see this in the drawing? (Lay the two distances down flat.)

$a$ (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 $\left( {0,0} \right)$ 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 $\displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$. For vertical ellipses, see the table below.

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 $c$. The foci are at $\left( \pm c,0 \right)$, 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:

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:

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

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, and find the sum of the distance from any point on the ellipse to the two foci:

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 $\displaystyle \frac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$. Plug in our values for $a,\,b,\,h,\,\,\text{and}\,k$, and we get $\displaystyle \frac{{{\left( x-4 \right)}^{2}}}{25}+\frac{{{\left( y+3 \right)}^{2}}}{4}=1$. Note that we didn’t have to have the coordinates of the foci to obtain the equation of the ellipse.

The constant sum for this ellipse is $2a=2\left( 5 \right)=10$.

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.

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; 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 absolute value of the differences of the distances from two fixed points (the foci) inside the hyperbola is always the same, $\left| {{{d}_{1}}-{{d}_{2}}} \right|=2a$.

This distance, $2a$, is called the focal radii distance, focal constant, or constant difference, with $a$ being the distance between the center of the hyperbola to a vertex; thus, $2a$ is the distance between the two vertices. Can you see this in the drawing? (Lay the two distances down flat.)

Note that the two parts of a hyperbola aren’t parabolas, and are called the branches of the hyperbola.

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 $\displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$. For vertical hyperbolas, the equations are $\displaystyle \frac{{{{y}^{2}}}}{{{{a}^{2}}}}-\frac{{{{x}^{2}}}}{{{{b}^{2}}}}=1$ and $\displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$.

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 $\left( \pm a,0 \right)$ for a hyperbola centered at the origin.

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 $b$); note that the co-vertices do not lie on the hyperbola; they are one what we call the central rectangle (or fundamental rectangle) of the hyperbola, whose diagonals are asymptotes for the hyperbola branches. The conjugate axis is along the $y$-axis for a horizontal hyperbola, and the co-vertices are at $\left( 0,\pm b \right)$ for a hyperbola centered at the origin.

The asymptotes for a horizontal hyperbola centered at $\left( {h,k} \right)$ are $\displaystyle y-k=\pm \frac{b}{a}\left( {x-h} \right)$. For a more generic equation, the asymptotes for a hyperbola centered at $(h,k)$ are $\displaystyle y-k=\pm \frac{{\sqrt{{\text{number under the }y}}}}{{\sqrt{{\text{number under the }x}}}}\left( {x-h} \right)$ (note that $\displaystyle \pm \frac{{\sqrt{{\text{number under the }y}}}}{{\sqrt{{\text{number under the }x}}}}$ are the slopes of the diagonals of the central rectangle of the hyperbola) – this works for both horizontal and vertical hyperbolas!

The focuses or foci always lie inside the curves on the major axis, and the distance from the center to a focus is $c$. 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 a hyperbole 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:

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:

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

Writing Equations of Hyperbolas

You may be asked to write an equation from either a graph or a description of a hyperbola, as in the following problem:

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 co-vertices are $\left( {-2,8} \right)$ and $\left( {-2,-2} \right)$. From this information, we can get the center (midpoint between the co-vertices), 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  $\displaystyle \frac{{{\left( x+2 \right)}^{2}}}{9}-\frac{{{\left( y-3 \right)}^{2}}}{25}=1$.

Problem:

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

Solution:

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

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 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 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:

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 $\displaystyle \frac{{{x}^{2}}}{1024}-\frac{{{y}^{2}}}{729}=1$. Thus, $a=\sqrt{{1024}}=32$. 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:

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

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

Solution:

Draw a picture first and make the nucleus the center of the hyperbola at $\left( {0,0} \right)$.

Problem:

When flying faster than the speed of sound, an airplane that flies parallel to the ground forms sound waves in the shape of a cone behind it. The waves intersect the ground in the shape of a hyperbola, with the airplane directly above the center of this hyperbola.

If you hear a sonic boom along the hyperbola $\displaystyle \frac{{{{x}^{2}}}}{{100}}-\frac{{{{y}^{2}}}}{4}=1$, what is the shortest horizontal distance you could be to the airplane?

Solution:

Draw a picture first and make the plane directly over the point $\left( {0,0} \right)$.

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 (right-hand side positive):  circle
• ${{x}^{2}}$ and ${{y}^{2}}$ with different coefficients  and  +  sign (right-hand side positive):  ellipse
• ${{x}^{2}}$ and ${{y}^{2}}$ with same coefficients and  –  sign (right-hand side positive):   hyperbola
• ${{x}^{2}}$ and ${{y}^{2}}$ with different coefficients and  –  sign (right-hand side positive):   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!

For Practice: Use the Mathway widget below to try a problem. Click on Submit (the blue arrow to the right of the problem) and click on Find the Foci to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

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On to Systems of Non-Linear Equations  – you are ready!

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