Verified Solution • 2025–26 Session moderate

QJEEM17Ma50fd: JEE Main 2017 Mathematics - Class 11 JEEM17Ma50fd Solution: The locus of the point of intersection of the straight lines,…

The locus of the point of intersection of the straight lines,

tx −- 2y −- 3t = 0

x −- 2ty + 3 = 0 (t ∈\in R), is :

Multiple Choice Options

  • A an ellipse with eccentricity 25{2 \over {\sqrt 5 }}
  • B an ellipse with the length of major axis 6
  • C a hyperbola with eccentricity 5\sqrt 5
  • D a hyperbola with the length of conjugate axis 3

Step-by-Step Verified Solution

Here, tx −- 2y −- 3t = 0  &  x −- 2ty + 3 = 0

On solving, we get;

y = 6t2t2−2{{6t} \over {2{t^2} - 2}} = 3tt2−1{{3t} \over {{t^2} - 1}} & x = 3t2+3t2−1{{3{t^2} + 3} \over {{t^2} - 1}}

Put    t = tan θ\theta

∴\therefore    x = −- 3 sec 2 θ\theta   &  2y = 3 ( −- tan 2 θ\theta )

∵\because    sec22 θ\theta −- tan22 θ\theta = 1

⇒\Rightarrow     x29{{{x^2}} \over 9} −- y29/4{{{y^2}} \over {9/4}} = 1

which represents at hyperbola

∴\therefore    a2 = 9  &  b2 = 9/4

λ\lambda (T.A.) = 6; e2 = 1 + 9/49{{9/4} \over 9} = 1 + 14{1 \over 4} ⇒\Rightarrow e = 52{{\sqrt 5 } \over 2}

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