Hyperbola
Normals
IIT-JEE 2011 Paper 2
Grade 11
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 the hyperbola is:
(1) $\frac{\sqrt{5}}{2}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\sqrt{2}$
(4) $\sqrt{3}$
Step-by-Step Solution
Key Concept: Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$. At $(6, 3)$ passing through $(9, 0)$: $\frac{9a^2}{6} = a^2 + b^2 \Rightarrow b^2 = a^2/2$. Then $e^2 = 1 + b^2/a^2 = 3/2$, so $e = \sqrt{3/2} = \sqrt{6}/2$. Also verify $P$ is on hyperbola: $36/a^2 - 9/b^2 = 1 \Rightarrow 36/a^2 - 18/a^2 = 1 \Rightarrow 18/a^2 = 1 \Rightarrow a^2 = 18$, $b^2 = 9$. $e = \sqrt{1 + 9/18} = \sqrt{3/2} = \sqrt{6}/2$.
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Correct Answer: (1)