Q52 / 90 · JEE Main 2018 (16 Apr Online) · Mathematics

Let $P$ be a point on the parabola $x^{2} = 4 y$ . If the distance of $P$ from the center of the circle $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at $P$ is

Solution

Let $P (2 t, t^{2})$ .

Equation normal at $P$ to $x^{2} = 4 y$ be

$y - t^{2} = - \frac{1}{t} (x - 2 t)$ .

It passes a normal through $(- 3, 0)$ .

$\Rightarrow 0 - t^{2} = - \frac{1}{t} (- 3 - 2 t)$

$\Rightarrow t^{3} + 2 t + 3 = 0$

$\Rightarrow (t + 1) (t^{2} - t + 3) = 0$

$\Rightarrow t = - 1$

So, point $P$ is $(- 2, 1)$ .

Equation of a tangent to $x^{2} = 4 y$ at $(- 2, 1)$ .

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

$∴ x + y + 1 = 0$ .

Hence, the correct answer is option: 'A'.