# Find the equation of the hyperbola having its foci at

Question:

Find the equation of the hyperbola having its foci at $(0, \pm \sqrt{14})$ and passing through the point P(3, 4).

Solution:

Given: Foci at $(0, \pm \sqrt{14})$ and passing through the point $\mathrm{P}(3,4)$

Need to find: The equation of the hyperbola.

Let, the equation of the hyperbola be:

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

It passes through the point $P(3,4)$

So putting the values of $(x, y)$ we get,

$\frac{3^{2}}{a^{2}}-\frac{4^{2}}{b^{2}}=1 \Rightarrow \frac{9}{a^{2}}-\frac{16}{b^{2}}=1 \cdots(1)$

Foci at $(0, \pm \sqrt{14})$

So, ae $=\sqrt{14}$

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

$\Rightarrow a \sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{14}$

$\Rightarrow \sqrt{a^{2}+b^{2}}=\sqrt{14}$

$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=14$ [Squaring on both sides]

$\Rightarrow a^{2}=14-b^{2}$  .............(2)

Comparing (1) and (2) we get

$\frac{9}{14-b^{2}}-\frac{16}{b^{2}}=1$

$\frac{9}{14-b^{2}}=1+\frac{16}{b^{2}}=\frac{b^{2}+16}{b^{2}}$

$9 b^{2}=14 b^{2}-b^{4}+224-16 b^{2}$

$b^{4}+11 b^{2}-224=0$

Solving the equations we get,

$b_{1}=\sqrt{\frac{1}{2}(-11+3 \sqrt{113})}$

$b_{2}=-\sqrt{\frac{1}{2}(-11+3 \sqrt{113})}$

$b_{3}=(-i) \sqrt{\frac{1}{2}(11+3 \sqrt{113})}$

$b_{4}=i \sqrt{\frac{1}{2}(11+3 \sqrt{113})}$

With the help of any of these values of b we can't find out the equation of the hyperbola.

* This is the only process we can apply in this standard.
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