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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2020 · 6 September · Morning Shift · Mathematics · Parabola

Let L _ 1 be a tangent to the parabola y ^ 2 = 4 ( x + 1 ) and L _...

Let L1{L}_{1} be a tangent to the parabola y2=4(x+1){y}^{2} = 4 ( x + 1 ) and L2{L}_{2} be a tangent to the parabola y2=8(x+2){y}^{2} = 8 ( x + 2 ) such that L1{L}_{1} and L2{L}_{2} intersect at right angles. Then L1{L}_{1} and L2{L}_{2} meet on the straight line:

Options

  1. Ax + 3 = 0
  2. B2x + 1 = 0
  3. Cx + 2 = 0
  4. Dx + 2y = 0

Correct answer

A. x + 3 = 0

Step-by-step solution

Let tangent of y2=4(x+1){y}^{2} = 4 ( x + 1 ) be L1:t1y=(x+1)+t12{L}_{1}: {t}_{1}y = ( x + 1 ) + {t}_{1}^{2}
and tangent of y2=8(x+2){y}^{2} = 8 ( x + 2 ) be L2:t2y=(x+2)+2t22{L}_{2}: {t}_{2}y = ( x + 2 ) + 2{t}_{2}^{2}
L1⊥L2{L}_{1} ⊥ {L}_{2}
1t1⋅1t2=−1\dfrac{1}{{t}_{1}} \cdot \dfrac{1}{{t}_{2}} = - 1
t1t2=−1{t}_{1}{t}_{2} = - 1
t2×(i)−t1×(ii)⇒x+3=0{t}_{2} \times ( i ) - {t}_{1} \times ( ii ) ⇒ x + 3 = 0
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