Question

Let P(6, 3) be a point on hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ If the normal at the point P intersect the x - axis at (9, 0), then the eccentricity of the hyperbola is

(IIT JEE 2011)

• A. $\sqrt{\frac{5}{2}}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is B

$\sqrt{\frac{3}{2}}$

We have the equation of hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Equation of normal at point $(x_1,y_1)$ is
$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_2}=a^2+b^2$
Equation of normal at point (6, 3)
$\frac{a^{2}x}{6}-\frac{b^{2}y}{3}=a^2+b^2$
Equation of normal at point (9, 0)
$\frac{9a^2}{6}=a^2+b^2$
$\Rightarrow \frac{3a^2}{2}-a^2=b^2$
$\Rightarrow \frac{3a^2}{2a^2}-\frac{a^2}{a^2}=\frac{b^2}{a^2}$
$\Rightarrow \frac{3}{2}-1=\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$
$\Rightarrow a^2=2b^2$
Let the eccentricity of the hyperbola be e.
$e^2=1+\frac{b^2}{a^2}$
$e^2=1+\frac{1}{2}$
$\Rightarrow e^2=\frac{3}{2}$
$\Rightarrow e=\sqrt{\frac{3}{2}}$