Coordinate Geometry and Conics · Lesson 5

Ellipses and hyperbolas: center and axes from the equation

Ellipse. (x−h)2a2+(y−k)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 has center (h,k)(h, k). It reaches a units left and right of the center and b units up and down, so its horizontal axis has length 2a2a and its vertical axis has length 2b2b. The denominators are a2a^2 and b2b^2: take square roots before you measure, and double for a full axis.

Hyperbola. A minus sign between the squared terms makes a hyperbola. (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 opens left and right, with vertices a units left and right of the center (h,k)(h, k). When the y term is the positive one, (y−k)2a2−(x−h)2b2=1\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1, it opens up and down, with vertices a units above and below the center. The positive term sets the direction, not the larger denominator.

Naming a conic. Move every squared term to one side. In this course, a question that depends on the ellipse or hyperbola standard form states it in the stem; naming the conic you do from the signs.

Naming a conic from its squared terms
Squared terms (on one side)ConicExample
Both, same sign, equal coefficientsCirclex2+y2=9x^2 + y^2 = 9
Both, same sign, different coefficientsEllipse4x2+y2=164x^2 + y^2 = 16
Both, opposite signsHyperbolax2−y2=4x^2 - y^2 = 4
Only one variable squaredParabolay=x2−4y = x^2 - 4

Worked example Easy

An ellipse with equation (x−h)2a2+(y−k)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 has center (h,k)(h, k), a horizontal axis of length 2a2a and a vertical axis of length 2b2b. What is the length of the horizontal axis of the ellipse (x−2)225+(y+1)29=1\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1?

  1. 55
  2. 66
  3. 1010Answer
  4. 2525

How to solve it

  1. The x term's denominator is a2=25a^2 = 25, so a=5a = 5.
  2. The horizontal axis is 2a=2⋅5=102a = 2 \cdot 5 = 10.

Why each choice is right or wrong

  • A. Incorrect. 5 is a, the distance from the center to one end of the horizontal axis. The full axis is 2a2a.
  • B. Incorrect. 6 is 2b2b, the length of the vertical axis. The horizontal axis comes from the x term's denominator.
  • C. Correct. a2=25a^2 = 25 gives a=5a = 5, so the horizontal axis has length 2a=102a = 10.
  • D. Incorrect. 25 is a2a^2, the denominator, not a length. Take its square root and double it.

Worked example Medium

A hyperbola with equation (y−k)2a2−(x−h)2b2=1\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 has center (h,k)(h, k) and opens up and down, with vertices a units above and below the center. What are the vertices of the hyperbola (y−3)216−(x+1)24=1\frac{(y - 3)^2}{16} - \frac{(x + 1)^2}{4} = 1?

  1. (−1,−1)(-1, -1) and (−1,7)(-1, 7)Answer
  2. (−5,3)(-5, 3) and (3,3)(3, 3)
  3. (−1,1)(-1, 1) and (−1,5)(-1, 5)
  4. (−1,−13)(-1, -13) and (−1,19)(-1, 19)

How to solve it

  1. Center: (h,k)=(−1,3)(h, k) = (-1, 3).
  2. a2=16a^2 = 16, so a=4a = 4. The y term is positive, so the vertices are above and below the center.
  3. (−1,3−4)=(−1,−1)(-1, 3 - 4) = (-1, -1) and (−1,3+4)=(−1,7)(-1, 3 + 4) = (-1, 7).

Why each choice is right or wrong

  • A. Correct. The center is (−1,3)(-1, 3) and a=16=4a = \sqrt{16} = 4, so the vertices are 4 units below and above it: (−1,−1)(-1, -1) and (−1,7)(-1, 7).
  • B. Incorrect. (−5,3)(-5, 3) and (3,3)(3, 3) move 4 units left and right, as if the hyperbola opened sideways. The positive y term means it opens up and down.
  • C. Incorrect. (−1,1)(-1, 1) and (−1,5)(-1, 5) use 4=2\sqrt{4} = 2, from the x term. The distance a comes from the denominator under the positive y term, 16.
  • D. Incorrect. (−1,−13)(-1, -13) and (−1,19)(-1, 19) use a2=16a^2 = 16 as the distance. Take the square root: a=4a = 4.

Trap a, a squared, or 2a

In x236+y24=1\frac{x^2}{36} + \frac{y^2}{4} = 1, 36 is a2a^2, so a=6a = 6 and the horizontal axis is 2a=122a = 12. The usual slips: using 36, stopping at 6, or using the other denominator. For a hyperbola, the vertices are a units from the center, not a2a^2 and not 2a2a.

Calculator When a calculator is faster

Graphing is quick here: type the equation as given into the built-in Desmos calculator if you test online and read the center, the ends of the axes or the vertices off the grid. It also settles which way a hyperbola opens. A handheld calculator that graphs only y = … makes you solve for y first, so reading the standard form is faster there.