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 :

Engineering

Mathematics

Normal to Hyperbola

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 the point P intersects the x‑axis at (9, 0), then the eccentricity of the hyperbola is :

$\sqrt{\frac{5}{2}}$

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

$\sqrt{2}$

$\sqrt{3}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Equation of the normal at (6,3) is

$\frac{a^{2} x}{6} + \frac{b^{2} y}{3} = a^{2} + b^{2}$

It passes through (9, 0)

$so \frac{9 a^{2}}{6} = a^{2} + b^{2}$

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

$∴ e^{2} - 1 = \frac{1}{2}$

$e^{2} = \frac{3}{2} \Rightarrow e = \sqrt{\frac{3}{2}}$