Question

Find the eccentricity, coordinates of foci, equation of directrix and length of the latus-rectum of the hyperbola $2x^2-3y^2=5$

Answer 1

Equation of hyperbola $2x^2-3y^2=5$

$\Rightarrow \frac{x^2}{\frac{5}{2}}-\frac{y^2}{\frac{5}{3}}=1$

$\therefore a^2=\frac{5}{2}\text{and}{b}^2=\frac{5}{3}$

For eccentricity

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

$\Rightarrow e=\sqrt{1+\frac{\frac{5}{3}}{\frac{5}{2}}}$

$\Rightarrow e=\sqrt{1+\frac{2}{3}}$

$\Rightarrow e=\sqrt{\frac{5}{3}}$

For coordinate of focus

$\because a^2=\frac{5}{2}\Rightarrow a=\sqrt{\frac{5}{2}}$

$\therefore ae=\sqrt{\frac{5}{2}}\times \sqrt{\frac{5}{3}}$

$\Rightarrow ae=\frac{5}{\sqrt{6}}$

$\because S(ae,0)\text{and}S\text{'}(-ae,0)$

$\Rightarrow S(\frac{5}{\sqrt{6}},0)\text{and}S\text{'}(-\frac{5}{\sqrt{6}},0)$

Equation of directrix
$x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}}$

$\Rightarrow x=\pm \sqrt{\frac{3}{2}}$

$\Rightarrow \sqrt{2}x\mp \sqrt{3}=0$

Length of latus-rectum

$L.R.=\frac{2b^2}{a}$

$\Rightarrow L.R.=\frac{2\times \frac{5}{3}}{\sqrt{\frac{5}{2}}}$

$\Rightarrow L.R.=\frac{2}{3}\sqrt{10}$

Final answer:

$e=\sqrt{\frac{5}{3}}$, Foci $(\pm \frac{5}{\sqrt{6}},0)$, Equation of directrix $\sqrt{2}x\mp \sqrt{3}=0$ and $L.R.=\frac{2}{3}\sqrt{10}$