Question

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

Answer 1

Given eccentricity, $e=3$
Since, we know that $e=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow 3=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow 9=1+\frac{b^2}{a^2}$ [Squaring on both sides]
$\Rightarrow \frac{b^2}{a^2}=9-1$
$\Rightarrow \frac{b^2}{a^2}=8$
$\therefore b^2=8a^2$.........(i)
And also we have length of the latus rectum $=\frac{2b^2}{a}=4$
$\Rightarrow b^2=2a$........(ii)

From equation(i) and (ii), we have $8a^2=2a$
$8a^2-2a=0$
$\Rightarrow 2a(4a-1)=0$
$\Rightarrow 4a-1=0$
$\Rightarrow 4a=1$
$\Rightarrow a=\frac{1}{4}$ as $a\ne 0$

$\therefore a^2=\frac{1}{16}$ and $b^2=(2\times \frac{1}{4})=\frac{1}{2}.$
Hence, the equation of hyperbola is $\frac{x}{a^2}-\frac{y}{b^2}=1$.....(iii)

Substitute $a^2=\frac{1}{16}$ and $b^2=\frac{1}{2}$ in equation(iii), we get
$\frac{x}{\frac{1}{16}}-\frac{y}{\frac{1}{2}}=1$
$\therefore 16x^2-2y^2=1$ is the required hyperbola equation.