IIT-JEE 2011 · Paper 2 · Mathematics Q41

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MathematicsSingle correct+3 / -1Question 41 of 60

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

\sqrt{\dfrac52}

\sqrt{\dfrac32}

\sqrt2

\sqrt3

Answer and worked solution

Official answer: B

Normal at $(x_1,y_1)$ : $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$ . At $P(6,3)$ with $y=0$ :
$x=\frac{6(a^2+b^2)}{a^2}=9\implies\frac{b^2}{a^2}=\frac12.$

Eccentricity. $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac32}$ .

(B)

Topics:Hyperbola Tangents and normals to conics

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IIT-JEE 2011 · Paper 2, Mathematics question 41