Let P be the point on the parabola {y^2} = 8x | vedclass

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Question

(C) The minimum distance from a point to a curve occurs along the normal to the curve at that point. For the parabola ${y^2} = 8x$,we have $4a = 8$,so

Vedclass · Accepted Answer

${x^2} + {y^2} - 4x + 8y + 12 = 0$

Vedclass · Answer

(C) The minimum distance from a point to a curve occurs along the normal to the curve at that point. For the parabola ${y^2} = 8x$,we have $4a = 8$,so $a = 2$. The normal to the parabola at point $P(at^2, 2at) = (2t^2, 4t)$ is given by $y = -tx + 2at + at^3$. Substituting $a = 2$,the normal is $y = -tx + 4t + 2t^3$. Since this normal passes through the centre of the circle $C(0, -6)$,we have: $-6 = -t(0) + 4t + 2t^3 $ $\Rightarrow 2t^3 + 4t + 6 = 0 $ $\Rightarrow t^3 + 2t + 3 = 0$. By inspection,$t = -1$ is a root. For $t = -1$,the point $P$ is $(2(-1)^2, 4(-1)) = (2, -4)$. The distance $CP$ is the radius $r$ of the required circle: $r^2 = CP^2 = (2 - 0)^2 + (-4 - (-6))^2 = 2^2 + 2^2 = 4 + 4 = 8$. The equation of the circle with centre $P(2, -4)$ and radius squared $r^2 = 8$ is: $(x - 2)^2 + (y + 4)^2 = 8 $ $\Rightarrow x^2 - 4x + 4 + y^2 + 8y + 16 = 8 $ $\Rightarrow x^2 + y^2 - 4x + 8y + 12 = 0$.

Let $P$ be the point on the parabola ${y^2} = 8x$ which is at a minimum distance from the centre $C$ of the circle ${x^2} + {(y + 6)^2} = 1$. Then the equation of the circle,passing through $C$ and having its centre at $P$,is:

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