Question

Find the eqtion of the hyperbola whose vertices are $(0,\pm 3)$ and the eccetricity is $\frac{4}{3}$.Also find the coordinates of its foci.

Answer 1

The vertices of hyperbola are $(0,\pm a)$ and given vertices are $(0,\pm 3)$.

$\therefore$ $a=3$

Here, the given hyperbola is vertical hyperbola

So, required equation will be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$\Rightarrow$ $e=\frac{4}{3}$

$\Rightarrow$ $c=ae=3\times \frac{4}{3}=4$

$\Rightarrow$ Co-ordinates of foci $=(0,\pm c)=(0,\pm 4)$

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

$\Rightarrow$ $b^2=c^2-a^2=4^2-3^2$

$\Rightarrow$ $b^2=16-9=7$

$\therefore$ The equation of hyperbola is, $\frac{y^2}{3^2}-\frac{x^2}{7}=1$

$\Rightarrow$ $\frac{y^2}{9}-\frac{x^2}{7}=1$