Question

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 P intersects the x –axis at (9,0),then the eccentricity of its conjugate hyperbola is

• A. $\sqrt{\frac{2}{3}}$
  
• B. $\sqrt{\frac{3}{2}}$
  
• C. $\sqrt{\frac{1}{\sqrt{3}}}$
  
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

$\frac{dy}{dx}=\frac{x{b}^2}{y{a}^2}$
$\Rightarrow$ slope of normal at (6,3) =$\frac{-a^2}{2b^2}$
If passes through (9, 0)
$\Rightarrow \frac{-a^2}{2b^2}=1\Rightarrow e_1=\sqrt{\frac{3}{2}}\frac{1}{e_1^2}+\frac{1}{e_2^2}=1\Rightarrow e_2=\sqrt{3}$