If P(h, k) is a point on the parabola x = 4y^ | vedclass

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(D) The parabola is $x = 4y^2$,which can be written as $y^2 = \frac{1}{4}x$. Here $4a = \frac{1}{4}$,so $a = \frac{1}{16}$.Any point on the parabola i

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(D) The parabola is $x = 4y^2$,which can be written as $y^2 = \frac{1}{4}x$. Here $4a = \frac{1}{4}$,so $a = \frac{1}{16}$.Any point on the parabola is $P(at^2, 2at) = (\frac{t^2}{16}, \frac{t}{8})$.The normal at $P$ is $y = -tx + 2at + at^3$.Since the normal passes through $Q(0, 33)$,we have $33 = 2a(t) + a(t^3) = \frac{2t}{16} + \frac{t^3}{16}$.$528 = 2t + t^3 \Rightarrow t^3 + 2t - 528 = 0$.By inspection,$t = 8$ is a root: $512 + 16 - 528 = 0$.Thus,$P = (\frac{8^2}{16}, \frac{8}{8}) = (4, 1)$.The second parabola is $y^2 - 4y = 4x \Rightarrow (y - 2)^2 = 4(x + 1)$.This is a parabola with vertex $(-1, 2)$ and $4a = 4$,so $a = 1$.The directrix is given by $X = -a$,where $X = x + 1$.$x + 1 = -1 \Rightarrow x = -2$.The distance of $P(4, 1)$ from the line $x = -2$ is $|4 - (-2)| = 6$.

If $P(h, k)$ is a point on the parabola $x = 4y^2$ which is nearest to the point $Q(0, 33)$,then the distance of $P$ from the directrix of the parabola $y^2 = 4(x + y)$ is equal to:

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