# High School: Geometry

### Expressing Geometric Properties with Equations HSG-GPE.A.3

3. Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant.

For many students, solving problems in Geometryland can sometimes feel like a business trip to Algebraland. (It isn't a leisurely vacation, that's for sure!) Working with the equations of ellipses and hyperbolas requires fluency in two dialects of Mathese: both Geospeak and Algetongue.

Focus, vertex, asymptote, transverse, conjugate, standard position, Pythagorean theorem—all important geometric terms that students need to know. On the other hand, square, simplify, solve for x, and complete the square are crucial algebraic skills they need to be able to perform.

If their trip to Geometryland proves successful, they should know the equations that describe both ellipses (which look like elongated circles) and hyperbolas (which look like a pair of parabolas). For ellipses or hyperbolas with center (h, k), a major axis of a, and a minor axis of b, the formulas are:

Students should know that when the major axes are vertical rather than horizontal, the a2 and b2 values are switched. Whichever axis is elongated gets the larger a underneath it.

Chances are students will also want to know the distance formula. After all, the standard itself features the word "distances." Lucky for them, the distance formula is still the same as it always was.

Students should understand that while parabolas have a single focus, ellipses and hyperbolas have two of them. (The plural of focus is "foci" so don't let that trip them up.) The distance from one focus to any point on the ellipse and back to the other focus will always be constant. The same applies to hyperbolas, only instead of adding the distances, we subtract them.

Given a set of two foci, students should be able to find the equation of an ellipse. They can do this by setting up the equation in the form of two distances calculations (the distance from one focus to (x, y) and the other focus to that same point) that equal a constant, or by finding the values of h, k, a, and b.

It also helps to imagine a right triangle inside the ellipse with vertices at the center, one of the foci, and at the vertical vertex. The length of the hypotenuse is half the constant distance c (the other half is the hypotenuse of the triangle on the other side), the length of the base helps find the location of the foci, and the length of the vertical leg is b (the same b as in the equations above). Students can also find a by dividing the constant distance by 2.

You might consider representing the constant distance as a string loosely connected to two foci. Using the string to guide your marker, draw the ellipse and explain the relationship between the values and distances. Bringing in triangles will help mathematically cement these relationships. Students should familiarize themselves with all the intricacies of Ellipseville so that they can actually get around (and around and around).

The itinerary for Hyperbola City should look about the same. Students should be able to perform essentially the same calculations, finding the equation when given the foci. The only difference is that instead of a sum of two distances, we're talking about a difference.

If students are struggling, they might benefit from:

• reviewing the algebra used to simplify big equations, especially completing the square
• considering how the Pythagorean Theorem is applied here
• focusing on when and where to use addition and subtraction
• referring to a labeled drawing of the shape in question

#### Drills

1. What is the equation for an ellipse with foci (±2, 0) and constant distance 5?

With c = 2 and , we find that  and . The ellipse's vertex is on the origin, so h and k are both 0. (B) is incorrect because it places the vertex on (2, 0) and has incorrect values for a2 and b2, while (C) places the vertex on (-2, 0) as well as fails to square the a and b terms. Answer choice (D) incorrectly used b2 and c2 in place of a2 and b2.

2. What is the equation for an ellipse with one focus at (5, 8), center at (9, 8), and constant distance 16?

The vertex is given in the problem, as is the distance 2a = 16, and we can find that c = 4. This gives us a2 = 64 and b2 = 48. Each of the other choices includes a different point as the vertex (one of the foci or a point between the given focus and the actual vertex) and has incorrect values for a2 and b2.

3. What is the equation for an ellipse with one focus at (-2, -5), center at (-2, 1), and constant distance 18?

From the information given, we can find that a2 = 81 and b2 = 45. Because the focus and the vertex share an x coordinate, we also know that this ellipse is vertically oriented, meaning the term under the y side of the equation will be larger than that under the x. Both (B) and (D) share the same values, but they are backwards in (B). Both (A) and (C) have used the b and c terms rather than the a and b.

4. What is the equation for an ellipse with foci (±4, 2) and constant distance 12?

With the given foci, the vertex is at (0, 2) and c = 16; we are also given that 2a = 12. This gives us a2 = 36 and b2 = 20. Choice (A) does not square the a and b terms, and it lists the y coordinate of the vertex as -2. Choice (C) is incorrect because it yields a vertex (-4, 2) and uses incorrect terms for a and b. The vertex in (D) is (4, -2), which is also incorrect.

5. What is the equation for an ellipse with foci (1, 3) and (7, 3) and constant distance 8?

This ellipse has a vertex at (4, 3) and we are given that 2a = 8. We can calculate that a2 = 16, b2 = 7, and c2 = 9. The vertex of (A) is given as (-3, -4) and has a and b raised to their fourth powers. Choice (B) has the vertex on the origin and hasn't squared the a and b terms. Choice (C) includes the coordinates of the foci as the origin.

6. Which of these best describes the shape of the ellipse ?

An ellipse whose foci are at its center

The ellipse described is actually a circle with vertex at (-3, -4), which is also the location of its two foci. The fact that a2 = b2 is the giveaway that this is a special case ellipse. To prove this, we can multiply both sides of the equation by 9 to get (x + 3)2 + (y + 4)2 = 9, the equation of a circle with radius 3.

7. What is the equation for a hyperbola with foci (±5, 0) and transverse axis length of 8?

With foci at (±5, 0), we are given that c = 5 and we can find that the center is on the origin, so h = k = 0. Half the transverse axis length gives a = 4. Using c2 = a2 + b2, or 25 = 16 + b2, we find that b2 = 9. The foci lie on the x-axis, so the x term will be positive. Because the transverse lies on the x-axis, we use a2 as the denominator on the x term. This leaves the y term to be negative with the denominator of b2.

8. What is the equation for a hyperbola with foci (±13, 0) and transverse axis length of 10?

Hopefully you can tell that (D) is not only wrong, but dangerously wrong. The foci are given at (±13, 0), meaning c = 13, and the vertex is at the origin, so h = k = 0. The transverse axis length of 10 = 2a, so a = 5. Using c2 = a2 + b2, we find that b2 = 25. The foci lie on the x-axis, so the x term will be positive. Because the transverse lies on the x-axis, we use a2 as the denominator on the x term. This leaves the y term to be negative with the denominator of b2.

9. What is the equation for a hyperbola with foci (0, ±10) and transverse axis length of 16?

From the given information, we can find that our center is on the origin, that c=10, and that a=8. If we use c2 = a2 + b2, we can calculate that b = 6. Because the foci are on the y-axis, the y term will be positive. Because the transverse lies on the y-axis, we use a2 as the denominator on the y term. This leaves the x term to be negative with the denominator of b2. Since it's centered on the origin, h = k = 0.

10. Which description best matches the hyperbola with the following equation?