If P(h,k) is a point on the parabola x=4y^2 which is nearest to the point Q(0,33), then th…

Question

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:

Accepted Answer

Write $x=4y^2$ as $y^2=\dfrac{x}{4}$, i.e. $y^2=4ax$ with $a=\dfrac1{16}$. The normal at parameter $t$ (point $(at^2,2at)$) is $y=-tx+2at+at^3$. Through $Q(0,33)$:
$$33=\frac{t}8+\frac{t^3}{16}\ \Longrightarrow\ t^3+2t-528=0\ \Longrightarrow\ (t-8)(t^2+8t+66)=0\ \Longrightarrow\ t=8.$$

$$P=(at^2,2at)=\left(\frac{64}{16},\frac{16}{16}\right)=(4,1).$$

The second parabola: $y^2-4y=4x\Rightarrow(y-2)^2=4(x+1)$, vertex $(-1,2)$, directrix $x+1=-1\Rightarrow x=-2$.

Distance of $P(4,1)$ from $x=-2$ is $4-(-2)=6.$

Solutions · 1