Question

Find the equation of hyperbolas satisfying the following:
Vertices at $(0,\pm 6)$ and eccentricity$=\frac{5}{3}$

Answer 1

Here, the vertices are on x-axis

Therefore, the equation of hyperbola is in the form

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

Now, coordinates of vertices are $(0,\pm a)$ & given verices $=(0,\pm 6)$

So, $a=6$

We know that

Eccentricity$=e=\frac{c}{a}$

$\frac{5}{3}=\frac{c}{a}$

$5a=3c$

$5\times 6=3c$

$c=\frac{30}{3}=10$

Also, we know

$c^2=a^2+b^2$

$100=36+b^2$

$100-36=b^2$

$b^2=64$

Putting values in hyperbola equation

$\frac{x^2}{36}-\frac{y^2}{64}=1$

Required equation of hyperbola