Question

The equation of a hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . If $P\left(\sqrt{2},\sqrt{5}\right)$ is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is $10$ , then its eccentricity is

• A. $\frac{3}{2}$
• B. $\frac{5}{4}$
• C. $\frac{16}{9}$
• D. $\frac{4}{3}$

Answer 1

Answer: (B)

$\because P$ lies on director circle $x^2+y^2=a^2-b^2$
$\therefore$ $a^2-b^2=7$ …(i)
Also, $2ae=10\Rightarrow ae=5$ …(ii)
Now, $e^2=\frac{a^2+b^2}{a^2}\Rightarrow a^2+b^2=25$ …(iii)
From (i) & (iii), we get,
$a^2=16,b^2=9$
$e=\sqrt{\frac{25}{16}}=\frac{5}{4}$