Question

In the hyperbola $4x^2-9y^2=36$, find the axes, the coordinates of the foci, the eccentricity, and the latus rectum.

Answer 1

Standard equation of hyperbola is

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

$4x^2-9y^2=36$

Dividing the given equation by $36$, we get

$\frac{x^2}{3^2}-\frac{y^2}{2^2}=1$

Comparing with standard equation, we get

$a=3b=2$

$\Rightarrow 2a=2\left(3\right)=6=$ Transverse axes

and conjugate axis is $2b=2\left(2\right)=4$

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

$\Rightarrow e=\sqrt{1+\frac{4}{9}}$

$\Rightarrow e=\sqrt{\frac{13}{9}}$

Coordinates of foci

$S=(ae,0)S^{{}^{\prime }}=(-ae.0)$

$S=(3\times \sqrt{\frac{13}{9}},0)$

$S^{{}^{\prime }}=(-3\times \sqrt{\frac{13}{9}},0)$

$\Rightarrow (\pm \sqrt{13},0)$ are the coordinates of foci

Length of latus rectum is $\frac{2b^2}{a}=\frac{8}{3}$