Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ± $\frac{5}{6} x$

Answer 1

Answer:

$\frac{y^2}{100} -\frac{x^2}{144} =1$

Step-by-step explanation:

By looking at the characteristics given, it can be determined that the equation will be vertical.

Being that the center is the midpoint of the vertices, the center is (0,0).

We have determined that the equation is vertical, so the equation used to find the asymptotes is $y= k+\frac{a}{b} (x-h)\\ \\y=k-\frac{a}{b} (x-h)$ . ± $\frac{5}{6}$ x is the asymptote.

We know that a is equal to 10 because of the vertices, so a squared is equal to 100.

$\frac{a}{b} =\frac{5}{6}$

$\frac{5}{6} =\frac{10}{b}$

When you cross multiply, and solve for b, you should get 12. Because b is 12, b squared is 144.

Now just fill the vertical transverse axis equation.

$\frac{y^2}{100} -\frac{x^2}{144} =1$