Hyperbola Ques 31
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
(2011)
(a) $\sqrt{\frac{5}{2}}$
(b) $\sqrt{\frac{3}{2}}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$
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Answer:
Correct Answer: 31.(b)
Solution:
Formula:
- Equation of normal to hyperbola at $\left(x _1, y _1\right)$ is
$ \begin{aligned} & \quad \frac{a^{2} x}{x _1}+\frac{b^{2} y}{y _1}=\left(a^{2}+b^{2}\right) \\ & \therefore \quad \operatorname{At}(6,3)=\frac{a^{2} x}{6}+\frac{b^{2} y}{3}=\left(a^{2}+b^{2}\right) \\ & \because \text { It passes through }(9,0) . \Rightarrow \frac{a^{2} \cdot 9}{6}=a^{2}+b^{2} \\ & \Rightarrow \quad \frac{3 a^{2}}{2}-a^{2}=b^{2} \quad \Rightarrow \quad \frac{a^{2}}{b^{2}}=2 \\ & \therefore \quad e^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}} \end{aligned} $
BETA
