JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Parabola

The area in sq. units) of the smaller of the two circles that touch the parabola,

y^{2} = 4 x

at the points

(1, 2)

and the axis is

(A) $4 π (2 - 2)$
(B) $8 π (3 - 22)$ ✓ Correct
(C) $4 π (3 + 2)$
(D) $8 π (2 - 2)$

Answer: (B)

Solution

The tangent to

y^{2} = 4 x

(1, 2)

y = x + 1

, and the normal there is

y = - x + 3

; the touching circle's centre lies on this normal, say at

(3 - r, r)

at distance

r

from the point of contact. Setting that distance equal to

r

(3 - r - 1)^{2} + (r - 2)^{2} = r^{2}

, i.e.

2 (2 - r)^{2} = r^{2}

, which gives

r^{2} - 8 r + 8 = 0

and

r = 4 \pm 22

. The smaller circle takes

r = 4 - 22

, so its area is

π r^{2} = π (24 - 162) = 8 π (3 - 22)

square units.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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