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JEE Main 202116 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2021 · 16 March · Evening Shift · Mathematics · Parabola

Let C be the locus of the mirror image of a point on the parabola y...

Let CC be the locus of the mirror image of a point on the parabola y2=4x{y}^{2} = 4x with respect to the line y=xy = x. Then the equation of tangent to CC at $P\left(2, 1\right) is :

Options

  1. Ax−y=1x - y = 1
  2. B2x+y=52x + y = 5
  3. Cx+3y=5x + 3y = 5
  4. Dx+2y=4x + 2y = 4

Correct answer

A. x−y=1x - y = 1

Step-by-step solution

Given y2=4x{y}^{2} = 4x

Mirror image on y=x⇒C:x2=4yy = x ⇒ C: {x}^{2} = 4y

2x=4⋅dydx⇒dydx=x22x = 4 \cdot \dfrac{dy}{dx} ⇒ \dfrac{dy}{dx} = \dfrac{x}{2}

dydxP(2,1)=22=1{\dfrac{dy}{dx}}_{P\left(2, 1\right)} = \dfrac{2}{2} = 1

Equation of tangent at \left(2, 1\right)</p><p></p><p>⇒ y - 1 = 1 \times \left(x - 2\right)

⇒x−y=1⇒ x - y = 1

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