Let P be a point on the parabola x^2 = 4y. If | vedclass

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(C) Let the point $P$ on the parabola $x^2 = 4y$ be $(2t, t^2)$.The given circle is $x^2 + y^2 + 6x + 8 = 0$. Its centre $C$ is $(-3, 0)$.For the dist

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$x + y + 1 = 0$

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(C) Let the point $P$ on the parabola $x^2 = 4y$ be $(2t, t^2)$.The given circle is $x^2 + y^2 + 6x + 8 = 0$. Its centre $C$ is $(-3, 0)$.For the distance $PC$ to be minimum,the line $PC$ must be the normal to the parabola at $P$.The slope of the tangent to the parabola at $P(2t, t^2)$ is $\frac{dy}{dx} = \frac{x}{2} = t$.Therefore,the slope of the normal at $P$ is $-\frac{1}{t}$.The slope of the line $PC$ connecting $P(2t, t^2)$ and $C(-3, 0)$ is $\frac{t^2 - 0}{2t - (-3)} = \frac{t^2}{2t + 3}$.Equating the slopes: $\frac{t^2}{2t + 3} = -\frac{1}{t}$.$t^3 = -2t - 3 \Rightarrow t^3 + 2t + 3 = 0$.By inspection,$t = -1$ is a root: $(-1)^3 + 2(-1) + 3 = -1 - 2 + 3 = 0$.For $t = -1$,the point $P$ is $(2(-1), (-1)^2) = (-2, 1)$.The slope of the tangent at $P$ is $t = -1$.The equation of the tangent at $(-2, 1)$ is $y - 1 = -1(x + 2)$.$y - 1 = -x - 2 \Rightarrow x + y + 1 = 0$.

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:

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