Discussion Forum

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

Answer:

Explanation:

Equation of normal to hyperbola at $(x_{1},y_{1})$ is

 $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2}$

 $\therefore$  at (6,3)   , $\frac{a^{2}x}{6}+\frac{b^{2}y}{3}=a^{2}+b^{2}$

   It passes throught (9.0)

Now, $\frac{a^{2}.9}{6}=a^{2}+b^{2}$

 $\Rightarrow$  $\frac{3a^{2}}{2}-a^{2}=b^{2}\Rightarrow \frac{a^{2}}{b^{2}}=2$

 $\therefore$  $e^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{1}{2} \Rightarrow  e=\sqrt{\frac{3}{2}}$