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

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

The tangent to the parabola y ^ 2 = 4x at the point where it...

The tangent to the parabola y2=4x{y}^{2} = 4x at the point where it intersects the circle x2+y2=5{x}^{2} + {y}^{2} = 5 in the first quadrant, passes through the point:The tangent to the parabola y ^ 2 = 4x at the point where it intersects the circle x ^ 2 + y ^ 2 = 5 in the first

Options

  1. A(14,34)\left(\dfrac{1}{4}, \dfrac{3}{4}\right)
  2. B(−13,43)\left(- \dfrac{1}{3}, \dfrac{4}{3}\right)
  3. C(−14,12)\left(- \dfrac{1}{4}, \dfrac{1}{2}\right)
  4. D(34,74)\left(\dfrac{3}{4}, \dfrac{7}{4}\right)

Correct answer

D. (34,74)\left(\dfrac{3}{4}, \dfrac{7}{4}\right)

Step-by-step solution

The tangent to the parabola y ^ 2 = 4x at the point where it intersects the circle x ^ 2 + y ^ 2 = 5 in the first
Let’s first find points of intersection of two curves.
x2+4x=5{x}^{2} + 4x = 5
⇒x=1,−5(⇒ x = 1, - 5 ( neglect x=−5x = - 5 )
Hence point of intersections are \left(1, 2\right) and \left(1, - 2\right) .
Tangent to the parabola at first quadrant point \left(1, 2\right)
y⋅2=2(x+1)y ⋅ 2 = 2\left(x + 1\right)
⇒y=x+1⇒ y = x + 1 clearly point \left(\dfrac{3}{4}, \dfrac{7}{4}\right) satisfy the given equation of tangent.
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