Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $5y^ 2 - 9x^ 2 = 36$

Answer 1

The given equation is $5y^ 2 - 9x ^2 = 36$

$⇒ \frac{5y^2}{36} –\frac{ 9x^2}{36} = \frac{36}{36}$

$\frac{y^2}{(\frac{36}{5})} – \frac{x^2}{4} = 1.......(1)$

On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{y^2}{a^2} –\frac{ x^2}{b^2} = 1$, we obtain $a= \frac{6}{\sqrt5}$ and b = 2

We know that $a ^2 + b^ 2 = c^ 2 .$

$∴ c^2 = \frac{36}{5} + 4$

$c^2 = \frac{56}{5}$

$c = \sqrt{(\frac{56}{5})}$

$= \frac{2\sqrt{14}}{\sqrt{5}}$

Therefore,
The coordinates of the foci are (0, $\frac{2\sqrt{14}}{\sqrt{5}}$) and (0, – $\frac{2\sqrt{14}}{\sqrt{5}}$).

The coordinates of the vertices are (0, $\frac{6}{\sqrt5}$) and (0, -$\frac{6}{\sqrt5}$).

Eccentricity, $e = \frac{c}{a} =\frac{ (\frac{2\sqrt{14}}{\sqrt{5}}) }{ (\frac{6}{\sqrt5}) }= \frac{\sqrt{14}}{3}$

Length of latus rectum = $\frac{2b^2}{a} =\frac{(2 \times 22)}{\frac{6}{\sqrt{5}}} = \frac{(2\times4)}{\frac{6}{\sqrt{5 }}}= \frac{4\sqrt5}{3}$