Find the equation of the hyperbola whose foci are
Question:

Find the equation of the hyperbola whose foci are $(\pm \sqrt{29}, 0)$ and the transverse axis is of the length 10 .

Solution:

Given: Foci are $(\pm \sqrt{29}, 0)$, the transverse axis is of the length 10

Need to find: The equation of the hyperbola.

Let, the equation of the hyperbola be:

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

The transverse axis is of the length 10, i.e., $2 \mathrm{a}=10$

Therefore, $a=5$

The foci are given at $(\pm \sqrt{29}, 0)$ That means, ae $=\pm \sqrt{29}$, where e is the eccentricity.

$\Rightarrow 5 \mathrm{e}=\sqrt{29}[$ As $\mathrm{a}=5]$

$\Rightarrow \quad e=\frac{\sqrt{29}}{5}$

We know that,

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

Therefore,

$\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{\sqrt{29}}{5}$

$\Rightarrow 1+\frac{b^{2}}{a^{2}}=\frac{29}{25}$ [Squaring both sides]

$\Rightarrow \frac{b^{2}}{a^{2}}=\frac{29}{25}-1=\frac{4}{25}$

$\Rightarrow \mathrm{b}^{2}=\mathrm{a}^{2} \frac{4}{25}$

$\Rightarrow \mathrm{b}^{2}=25 \times \frac{4}{25}=4[$ As $\mathrm{a}=5]$

So, the equation of the hyperbola is,

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{25}-\frac{y^{2}}{4}=1$ [Answer]