# Find the equation of the hyperbola whose vertices are (0, ±3) and the

Question:

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

Solution:

Given: Vertices are $(0, \pm 3)$ and the eccentricity is $\frac{4}{3}$

Need to find: The equation of the hyperbola and coordinates of foci.

Let, the equation of the hyperbola be:

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

Vertices are $(\pm 3,0)$, that means, $a=3$

And also given, the eccentricity, $e=\frac{4}{3}$

We know that, $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$

Therefore,

$\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{4}{3}$

$\Rightarrow 1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{16}{9}$ [Squaring both sides]

$\Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{16}{9}-1=\frac{7}{9}$

$\Rightarrow \mathrm{b}^{2}=\frac{7}{9} \mathrm{a}^{2}=\frac{7}{9} \times 9=7[$ As $\mathrm{a}=3]$

So, the equation of the hyperbola is,

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

Coordinates of the foci $=(\pm a e, 0)=(\pm 4,0)$ [Answer]