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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Hyperbola

If the eccentricity of the standard hyperbola passing through the...

If the eccentricity of the standard hyperbola passing through the point (4,6)( 4,6 ) is 2,2, then the equation of the tangent to the hyperbola at (4,6)( 4,6 ) is:

Options

  1. A2x−3y+10=02x - 3y + 10 = 0
  2. Bx−2y+8=0x - 2y + 8 = 0
  3. C3x−2y=03x - 2y = 0
  4. D2x−y−2=02x - y - 2 = 0

Correct answer

D. 2x−y−2=02x - y - 2 = 0

Step-by-step solution

Let us suppose equation of hyperbola is
x2a2−y2b2=1\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1
e=2e = 2

∵b2=(e2−1)a2∵ {b}^{2} = \left({e}^{2} - 1\right){a}^{2}

⇒b2=3a2⇒ {b}^{2} = 3{a}^{2}
It is passing through \left(4,6\right)</p><p></p><p>\dfrac{16}{{a}^{2}} - \dfrac{36}{{b}^{2}} = 1</p><p></p><p>⇒ {a}^{2} = 4, {b}^{2} = 12<br/>So,equationoftangentis<br />So, equation of tangent isT = 0</p><p></p><p>\dfrac{x{x}_{1}}{{a}^{2}} - \dfrac{y{y}_{1}}{{b}^{2}} = 1</p><p>Tangentatpoint</p><p>Tangent at point( 4, 6 )is<br/>is<br />⇒ x - \dfrac{y}{2} = 1<br/><br />⇒ 2x - y - 2 = 0$

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