JEE Main 2019 — Parabola Question with Solution

Equation of tangent to the parabola y² = 4x at P(1, 2),

T = 0

2y = $$4\left(\frac{x+1}{2}\right)$$

$$\Rightarrow$$ y = x + 1

Equation of normal of the tangent at point P(1, 2)

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

$$\Rightarrow$$ y - 2 = - x + 1

$$\Rightarrow$$ x + y - 3 = 0

This normal also passes through the center (h, r) of the circle.

$$\therefore$$ h + k - 3 = 0

$$\Rightarrow$$ h = 3 - r

So center is (3 - r, r)

From picture you can see,

PC = r

$$\Rightarrow$$ (PC)² = r²

$$\Rightarrow$$ (3 - r - 1)² + (r - 2)² = r²

$$\Rightarrow$$ 4 + r² - 4r + r² + 4 - 4r = r²

$$\Rightarrow$$ r² - 8r + 8 = 0

$$\therefore$$ r = $$\frac{8\pm \sqrt{64-32}}{2}$$

$$\Rightarrow$$ r = $$\frac{8\pm \sqrt{32}}{2}$$

$$\Rightarrow$$ r = $$\frac{8\pm 4\sqrt{2}}{2}$$

$$\therefore$$ r = 4 + $$2\sqrt{2}$$ and 4 - $$2\sqrt{2}$$

If r = 4 + $$2\sqrt{2}$$ then center of the circle is

(-1 - $$2\sqrt{2}$$, 4 + $$2\sqrt{2}$$). From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.

So possible value of radius r = 4 - $$2\sqrt{2}$$

Then area of the circle

= $$\pi$$r²

= $$\pi$$(4 - $$2\sqrt{2}$$)²

= $$\pi$$(16 + 8 - $$16\sqrt{2}$$)

= $$8\pi \left(3-2\sqrt{2}\right)$$