Question

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

• A. $(a)\sqrt\frac{5}{2}$
• B. $(b)\sqrt\frac{3}{2}$
• C. $(c)\sqrt2$
• D. $(d)\sqrt3$

Answer 1

The Correct Option is B

Equation of normal to hyperbola at $(x_1,y_1)$ is
$\, \, \, \, \, \, \, \, \, \, \, \frac{a^2x}{x_1}+\frac{b^2y}{y_1}=(a^2+b^2)$
$\therefore\, \, \, \, At(6,3)=\frac{a^2x}{6}+\frac{b^2y}{3}=(a^2+b^2)$
$\because\, \, \, \, $ It passes through $(9, 0).\Rightarrow \frac{a^2.9}{6}=a^2+b^2$
$\Rightarrow\, \, \, \, \frac{3a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2$
$\therefore \, \, \, \, \, \, \, \, \, e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt\frac{3}{2}$