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  4. Let P be a point on the parabola, x2 = 4y. If the distance of P from the centre of the circle, x2 + y2 + 6 x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :

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

JEE MainJEE Main 2018Application of Derivatives

Solution:

Let $P\left(2 t, t^{2}\right)$, equation normal at $P$ to $x^{2}=4 y$ be
$y-t^{2}=\frac{-1}{t}(x-2 t)$
It passes through $(-3,0)$.
$0-t^{2} =\frac{-1}{t}(-3-2 t)$
$t^{3}+2 t+3 =0 $
$(t+1)\left(t^{2}-t+3\right) =0 $
$t =-1$
So, point $P$ is $(-2,1)$.
Therefore, equation of tangent to $x^{2}=4 y$ at $(-2,1)$ is
$x(-2) =2(y+1) $
$x+y+1 =0$