Question

The foci of the hyperbola $2x^2-3y^2=5$ is

• A. $(\pm \frac{5}{\sqrt{6}},0)$
• B. $(\pm \frac{5}{6},0)$
• C. $(\pm \frac{\sqrt{5}}{6},0)$
• D. None of these

Answer 1

The correct option is A $(\pm \frac{5}{\sqrt{6}},0)$

Given, equation of hyperbola is $2x^2-3y^2=5$

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

Here, $a^2=\frac{5}{2},b^2=\frac{5}{3}$

Now, $b^2=a^2(e^2-1)$

$\Rightarrow \frac{5}{3}=\frac{5}{2}(e^2-1)$

$\Rightarrow e^2-1=\frac{2}{3}$

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

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

$\therefore$ The foci of hyperbola is $(\pm ae,0)$

$=(\pm \sqrt{\frac{5}{2}}\cdot \sqrt{\frac{5}{3}},0)$

$=(\pm \frac{5}{\sqrt{6}},0)$