Question

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

• A. $x+4y-2=0$
• B. $x+2y=0$
• C. $x+y+1=0$
• D. $x-y+3=0$

Answer 1

The correct option is C $x+y+1=0$

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

Centre of circle $=(-g,-f)=(-3,0)$

For the distance between point $P$ and center of the circle to be minimum, the line is drawn from the center of the circle to point $P$ must be normal to the parabola at $P$.

Slope of line joining centre of circle to point $P=\frac{y_2-y_1}{x_2-x_1}=\frac{t^2-0}{2t+3}$

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

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

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

$⟹t^3+2t+3=0$

$⟹(t+1)(t^2-t+1)=0$

Real roots of the above equation are

$t=-1$

Coordinate of $P=(2t,t^2)=(-2.1)$

Slope of tangent to parabola at $P=\frac{dy}{dx}=\frac{x}{2}=t=-1$

Therefore, the equation of tangent is

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

$⟹x+y+1=0$

Hence, the answer is an option (C).