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JEE Main 202226 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
26 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2022 · 26 June · Evening Shift · Mathematics · Hyperbola

The normal to the hyperbola x ^ 2 a ^ 2 - y ^ 2 9 = 1 at the point...

The normal to the hyperbola x2a2−y29=1\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{9} = 1 at the point (8,33)\left(8, 3\sqrt{3}\right) on it passes through the point

Options

  1. A(15,−23)\left(15, - 2\sqrt{3}\right)
  2. B(9,23)\left(9, 2\sqrt{3}\right)
  3. C(−1,93)\left(- 1, 9\sqrt{3}\right)
  4. D(−1,63)\left(- 1, 6\sqrt{3}\right)

Correct answer

C. (−1,93)\left(- 1, 9\sqrt{3}\right)

Step-by-step solution

Given,

hyperbola: x2a2−y29=1\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{9} = 1

it passes through \left(8, 3\sqrt{3}\right)</p><p></p><p>∵ \dfrac{64}{{a}^{2}} - \dfrac{27}{9} = 1 ⇒ {a}^{2} = 16</p><p>Now,equationofnormaltohyperbolainpointform</p><p></p><p>Now, equation of normal to hyperbola in point form</p><p>∵ \dfrac{16x}{8} + \dfrac{9y}{3\sqrt{3}} = 16 + 9</p><p></p><p>⇒ 2x + \sqrt{3}y = 25</p><p>(−1,93)</p><p>\left(- 1, 9\sqrt{3}\right) satisfies the equation.

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