JEE Main 2016 (Online) 9th AprilQuestion 1
Let $a$ and $b$ respectively be the semi-transverse and semi-conjugate axes of a standard hyperbola whose eccentricity satisfies the equation $9e^{2}-18e+5=0$. If $S(5,0)$ is a focus and $5x=9$ is the corresponding directrix of this hyperbola, then $a^{2}-b^{2}$ is equal to
Hyperbola
Solution
$9e^2-18e+5=0\Rightarrow e=\frac53$ (hyperbola). $ae=5\Rightarrow a=3$, $\frac ae=\frac95$ checks; $b^2=25-9=16$. $a^2-b^2=-7$.