Normal to hyperbola frac{x^{2}}{a^{2}}-frac{y^{2}}{9}=1 at point (8,3 √{3}) on it passes through p

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10-2. Parabola, Ellipse, Hyperbola

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The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8,3 \sqrt{3})$ on it passes through the point

A

$(15,-2 \sqrt{3})$

B

$(9,2 \sqrt{3})$

C

$(-1,9 \sqrt{3})$

D

$(-1,6 \sqrt{3})$

(JEE MAIN-2022)

Solution

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1:(8,3 \sqrt{3})$ lie on Hyperbola then

$\frac{64}{a^{2}}-\frac{27}{9}=1 \Rightarrow a^{2}=\frac{64}{4}=16$

equation of normal at $(8,3 \sqrt{3})$ :

$\frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9$

$2 x+\sqrt{3} y=25$

Standard 11

Mathematics

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