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JEE Main 201816 AprilOnline
Exam
JEE Main
Year
2018
Date
16 April
Mode
Online
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2018 · 16 April · Online · Mathematics · Straight Lines

The locus of the point of intersection of the lines 2 x - y + 4 2 k...

The locus of the point of intersection of the lines 2x−y+42k=0\sqrt{2}x - y + 4\sqrt{2}k = 0 and 2kx+ky−42=0\sqrt{2}kx + ky - 4\sqrt{2} = 0 (kk is any non-zero real parameter) is

Options

  1. Aan ellipse whose eccentricity is 13\dfrac{1}{\sqrt{3}}
  2. Ba hyperbola whose eccentricity is 3\sqrt{3}
  3. Ca hyperbola with length of its transverse axis 828\sqrt{2}
  4. Dan ellipse with length of its major axis 828\sqrt{2}

Correct answer

C. a hyperbola with length of its transverse axis 828\sqrt{2}

Step-by-step solution

Given equation of lines are 2x−y+42k=0...(1)<p>and\sqrt{2}x - y + 4\sqrt{2}k = 0 . . . \left(1\right)<p>and\sqrt{2}kx + ky - 4\sqrt{2} = 0 . . . \left(2\right)</p><p>Eliminating</p><p>Eliminatingkfrom(1)and(2),</p><p>from \left(1\right) and \left(2\right),</p><p>⇒ \left(\sqrt{2}x + y\right)\left(\dfrac{\sqrt{2}x - y}{- 4\sqrt{2}}\right) = 4\sqrt{2}</p><p></p><p>⇒ 2{x}^{2} - {y}^{2} = - 32</p><p></p><p>⇒ \dfrac{{y}^{2}}{32} - \dfrac{{x}^{2}}{16} = 1,whichisahyperbola.</p><p>, which is a hyperbola.</p><p>⇒ e = \sqrt{1 + \dfrac{16}{32}} = \sqrt{\dfrac{3}{2}} and length of transverse axis

=82= 8\sqrt{2}.

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