Find the equation of the hyperbola whose foci are

Solution:

Given: Foci are (0, ±13), the conjugate axis is of the length 24

Need to find: The equation of the hyperbola.

Let, the equation of the hyperbola be:

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

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

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

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