Question

Which standard form of the equation of the hyperbola has vertices at? Thank you!

Answer 1

Answer:

x²/25 - y²/144 = 1

Explanation:

In a hyperbola with an equation

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The vertices are (a, 0) and (-a, 0) and the asymptotes are y = ±(b/a)x

In this case, the vertices are (5, 0) and (-5, 0), so the value of a is 5.

And the asymptote is y = ±(12/5)x, so the value of b is 12.

Then, replacing a = 5 and b = 12, we get that the equation of the hyperbola is

$\begin{gathered} \frac{x^2}{5^2}-\frac{y^2}{12^2}=1 \\ \\ \frac{x^2}{25}-\frac{y^2}{144}=1 \end{gathered}$

Therefore, the answer is

x²/25 - y²/144 = 1