Medium MCQ +4 / -1 PYQ · JEE Mains 2022

The normal to the hyperbola

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$ at the point $\left( {8,3\sqrt 3 } \right)$ on it passes through the point :

  1. A $\left( {15, - 2\sqrt 3 } \right)$
  2. B $\left( {9,2\sqrt 3 } \right)$
  3. C $\left( { - 1,9\sqrt 3 } \right)$ Correct answer
  4. D $\left( { - 1,6\sqrt 3 } \right)$

Solution

<p>Given hyperbola : ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$</p> <p>$\because$ It passes through $(8,3\sqrt 3 )$</p> <p>$\because$ ${{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16$</p> <p>Now, equation of normal to hyperbola</p> <p>${{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 + 9$</p> <p>$\Rightarrow 2x + \sqrt 3 y = 25$ ...... (i)</p> <p>$\left( { - 1,9\sqrt 3 } \right)$ satisfies (i)</p>

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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