# Find the equation of the hyperbola, the length of whose latus rectum is 4

Question:

Find the equation of the hyperbola, the length of whose latus rectum is 4 and the eccentricity is 3.

Solution:

Given: The length of latus rectum is 4, and the eccentricity is 3

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 length of the latus rectum is 4 units.

Therefore, $\frac{2 b^{2}}{a}=4 \Rightarrow b^{2}=2 a^{-\cdots}$ (1)

And also given, the eccentricity, $e=3$

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

Therefore

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

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

$\Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=8$

$\Rightarrow \mathrm{b}^{2}=8 \mathrm{a}^{2}$

$\Rightarrow 2 \mathrm{a}=8 \mathrm{a}^{2}[$ From $(1)]$

$\Rightarrow \mathrm{a}=\frac{1}{4}$

Therefore,

$b^{2}=2 a=2 \times \frac{1}{4}=\frac{1}{2}$

So, the equation of the hyperbola is,

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{1 / 16}-\frac{y^{2}}{1 / 2}=1 \Rightarrow 16 x^{2}-2 y^{2}=1$ [Answer]