Let P(6,3) be a point on the hyperbola frac{x | vedclass

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Question

Equation of normal to the hyperbola is

$\frac{ x - x _1}{\frac{ x _1}{ a ^2}}=\frac{ y - y _1}{-\frac{ y _1}{b^2}}$

So, equation of normal to hy

Vedclass · Accepted Answer

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

Vedclass · Answer

Equation of normal to the hyperbola is

$\frac{ x - x _1}{\frac{ x _1}{ a ^2}}=\frac{ y - y _1}{-\frac{ y _1}{b^2}}$

So, equation of normal to hyperbola at $(6,3)$ is

$\frac{x-6}{\frac{6}{a^2}}=\frac{y-3}{-\frac{3}{b^2}}$

Since, it intersects x -axis at $(9,0)$

$	ext { So, } \frac{9-6}{\frac{6}{a^2}}=\frac{-3}{-\frac{3}{b^2}}$

$\Rightarrow a^2=2 b^2$

Eccentricity of hyperbola $=\sqrt{1+\frac{ b ^2}{ a ^2}}$

$=\sqrt{1+\frac{ b ^2}{2 b^2}}=\sqrt{\frac{3}{2}}$

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

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