JEE Advanced · Mathematics · 15. Hyperbola
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
- A
\(\sqrt{\frac{5}{2}}\)
- B
\(\sqrt{\frac{3}{2}}\)
- C
\(\sqrt{2}\)
- D
\(\sqrt{3}\)
Answer & Solution
Correct Answer
(B)
\(\sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
Equation of normal to hyperbola at \(\left(x_1, y_1\right)\) is
\[
\begin{gathered}
\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\
\therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2
\end{gathered}
\]
It passes throught \((9,0)\).
\[
\begin{aligned}
& \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\
& \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\
& \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}
\end{aligned}
\]
\[
\begin{gathered}
\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\
\therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2
\end{gathered}
\]
It passes throught \((9,0)\).
\[
\begin{aligned}
& \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\
& \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\
& \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}
\end{aligned}
\]
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