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,
$a^{2}=16,b^{2}=9$
$e=\sqrt{\frac{25}{16}}=\frac{5}{4}$