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10-2. Parabola, Ellipse, Hyperbola
hard
Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$, then the eccentricity of the hyperbola is
A
$\sqrt{\frac{5}{2}}$
B
$\sqrt{\frac{3}{2}}$
C
$\sqrt{2}$
D
$\sqrt{3}$
(IIT-2011)
Solution
Equation of normal to the hyperbola is
$\frac{ x – x _1}{\frac{ x _1}{ a ^2}}=\frac{ y – y _1}{-\frac{ y _1}{b^2}}$
So, equation of normal to hyperbola at $(6,3)$ is
$\frac{x-6}{\frac{6}{a^2}}=\frac{y-3}{-\frac{3}{b^2}}$
Since, it intersects x -axis at $(9,0)$
$\text { So, } \frac{9-6}{\frac{6}{a^2}}=\frac{-3}{-\frac{3}{b^2}}$
$\Rightarrow a^2=2 b^2$
Eccentricity of hyperbola $=\sqrt{1+\frac{ b ^2}{ a ^2}}$
$=\sqrt{1+\frac{ b ^2}{2 b^2}}=\sqrt{\frac{3}{2}}$
Standard 11
Mathematics
