JEE Main 2018 — Parabola Question with Solution

Let P(2t, t²) be any point on the parabola.

Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = $$\frac{y_2-y_1}{x_2-x_1}$$ = $$\frac{t^2-0}{2t+3}$$

Also, slope of tangent to parabola at P = $$\frac{dy}{dx}$$ = $$\frac{x}{2}$$ = t

$$\therefore$$ Slope of normal = $$\frac{-1}{t}$$

$$\therefore$$ $$\frac{t^2-0}{2t+3}$$ = $$\frac{-1}{t}$$

$$\Rightarrow$$ t³ + 2t + 3 = 0

$$\Rightarrow$$ (t+1) (t² $$-$$ t + 3) = 0

$$\therefore$$ Real roots of above equation is

t = $$-$$ 1

Coordinate of P = (2t, t²) = ($$-$$2, 1)

Slope of tangent to parabola at P = t = $$-$$ 1

Therefore, equation of tangent is :

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

$$\Rightarrow$$ x + y + 1 = 0