Standard form equation for a hyperbola

Question

Find the standard form equation for a hyperbola with vertices at (5,0) and (-5,0) that passes through the point (6,11).

Answer 1

See the video.

Answer 2

As the vertices are only affected by the x-values, the standard form for a hyperbola that opens sideways would be (x-h)²/a^{2 -} (y-k)²/b² = 1 where (h,k) is the center of the hyperbola, "a" is half the length of the major axis, and "b" is half the length of the minor axis.

Our center in this case is (h,k)=(0,0) because of our vertices (5,0) and (-5,0) in which (0,0) is directly in-between.

Using either one of the vertices, we can use it to determine the values of a² and b²:

(x-h)²/a^{2 -} (y-k)²/b² = 1

(5-0)²/a^{2 -} (0-0)²/b² = 1

5²/a² = 1

25 = a²

x²/25 - y²/b² = 1

Because we are given a point of (6,11), we can determine the value of b²:

x²/25 - y²/b² = 1

6²/25 - 11²/b² = 1

36/25 - 121/b² = 1

-121/b² = -11/25

121/b² = 11/25

3025 = 11b²

275 = b²

Thus, the equation for the hyperbola with vertices (±5,0) and passes through (6,11) is x²/25 - y²/275 = 1

Hope this helped!