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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2021 · 18 March · Evening Shift · Mathematics · Hyperbola

Consider a hyperbola H: x^ 2 - 2y^ 2 = 4 . Let the tangent at a...

Consider a hyperbola H:x2−2y2=4H: x^{2} - 2y^{2} = 4. Let the tangent at a point P(4,6)P ( 4, \sqrt{6} ) meet the xx-axis at QQ and latus rectum at R(x1,y1),x1>0R\left({x}_{1}, {y}_{1}\right), {x}_{1} > 0. If FF is a focus of HH which is nearer to the point PP, then the area of ΔQFR\Delta QFR& (in sq. units) is equal toConsider a hyperbola H: x^ 2 - 2y^ 2 = 4 . Let the tangent at a point P ( 4, 6 ) meet the x -axis at Q and latus rectumConsider a hyperbola H: x^ 2 - 2y^ 2 = 4 . Let the tangent at a point P ( 4, 6 ) meet the x -axis at Q and latus rectum

Options

  1. A464\sqrt{6}
  2. B6−1\sqrt{6} - 1
  3. C76−2\dfrac{7}{6} - 2
  4. D46−14\sqrt{6} - 1

Correct answer

C. 76−2\dfrac{7}{6} - 2

Step-by-step solution

Consider a hyperbola H: x^ 2 - 2y^ 2 = 4 . Let the tangent at a point P ( 4, 6 ) meet the x -axis at Q and latus rectum

x24−y22=1\dfrac{x2}{4} - \dfrac{y2}{2} = 1

e=1+b2a2=32e = \sqrt{1 + \dfrac{b2}{a2}} = \sqrt{\dfrac{3}{2}}

∴∴ Focus F(ae,0)⇒F(6,0)F ( ae, 0 ) ⇒ F ( \sqrt{6}, 0 )

equation of tangent at PP to the hyperbola is
2x−y6=22x - y\sqrt{6} = 2

tangent meet xx-axis at Q(1,0)Q ( 1, 0 )

& latus rectum x=6x = \sqrt{6} at R \left(\sqrt{6},26\dfrac{2}{6}\left(\sqrt{6} -1\right)\right)

∴∴ Area of ΔQFR={Δ}_{QFR} =12\dfrac{1}{2}(6−1)\left(\sqrt{6} -1\right)\cdot 26\dfrac{2}{6} \left(\sqrt{6} -1\right)</p><p>=</p><p>=\dfrac{7}{6}$−-2$& sq. units

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