JEE PYQ: Parabola Question 17
Question 17 - 2019 (09 Apr Shift 2)
The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^2 = 4x$ at the point $(1, 2)$ and the $x$-axis is:
(1) $8\pi(2 - \sqrt{2})$
(2) $4\pi(2 - \sqrt{2})$
(3) $4\pi(3 + \sqrt{2})$
(4) $8\pi(3 - 2\sqrt{2})$
Show Answer
Answer: (4)
Solution
Tangent at $(1,2)$: $y = x + 1$. Circle touches $x$-axis and has tangent $y = x + 1$ at $(1,2)$. Family of circles: $(x-1)^2 + (y-2)^2 + \lambda(x - y + 1) = 0$. Circle touches $x$-axis: $y$-coordinate of centre equals radius. Solving: $\lambda = 4 - 4\sqrt{2}$ (smaller). Radius $= 4 - 2\sqrt{2}$. Area $= \pi(4 - 2\sqrt{2})^2 = 8\pi(3 - 2\sqrt{2})$.
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