Question

Find the equation of the hyperbola whose foci are $(\pm 3\sqrt{5},0)$ and the length of the latus rectum is 8 units

Answer 1

Co-ordinates of are $(\pm c,0)$ and given foci are $(\pm 3\sqrt{5},0)$

$\therefore$ $c=3\sqrt{5}$

Since, foci is on x-axis

Hence, equation of hyperbola is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

We know,

$\Rightarrow$ $c^2=a^2+b^2$

$\Rightarrow$ $(3\sqrt{5}{)}^2=a^2+b^2$

$\therefore$ $a^2+b^2=45$ ---- ( 1 )

Latus rectum $=8$

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

$\Rightarrow$ $b^2=4a$ --- ( 2 )

Substituting ( 2 ) in ( 1 ) we get,

$\Rightarrow$ $a^2+4a=45$

$\Rightarrow$ $a^2+4a-45=0$

$\Rightarrow$ $a^2+9a-5a-45=0$

$\Rightarrow$ $a(a+9)-5(a+9)=0$

$\Rightarrow$ $(a+9)(a-5)=0$

So, $a=5$ or $a=-9$

Since, $a$ is distance, it cant be negative.

$\therefore$ $a=5$

$\Rightarrow$ $b^2=4a=4\left(5\right)=20$

Now,

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

Putting values,

$\Rightarrow$ $\frac{x^2}{5^2}-\frac{y^2}{20}=1$

$\Rightarrow$ $\frac{x^2}{25}-\frac{y^2}{20}=1$