JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Parabola

Let

C

be the locus of the mirror image of a point on the parabola

y^{2} = 4 x

with respect to the line

y = x

. Then the equation of tangent to

C

P (2, 1)

is :

(A) $x - y = 1$ ✓ Correct
(B) $2 x + y = 5$
(C) $x + 2 y = 4$
(D) $x + 3 y = 5$

Answer: (A)

Solution

A point of

y^{2} = 4 x

has the form

(t^{2}, 2 t)

, and reflection in

y = x

interchanges the coordinates, so the image is

(2 t, t^{2})

. Writing

x = 2 t

and

y = t^{2}

and eliminating

t

y = \frac{x ^{2}}{4}, so C : x^{2} = 4 y

The point

P (2, 1)

lies on

C

because

2^{2} = 4 (1)

. Differentiating

x^{2} = 4 y

gives

\frac{d y}{d x} = \frac{x}{2}

, which equals

1

x = 2

, so the tangent is

y - 1 = 1 \cdot (x - 2)

The tangent to

C

P

x - y = 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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