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# Finding distance to parabola's focus, given some points

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A high school student I know has the following problem:

A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-axis.

1. Find the distance of $D$ from the directrix of the parabola, in terms of $p$.

We draw two circles. One has its center at $D$ and a radius of $p+4$. The other has its center at the focus $F$ of the parabola. The latter circle is externally tangent to the first circle, and is tangent also to the $y$-axis.

1. Using part 1, find the equation of the parabola.

The student found $D=(\frac{32}p,8)$ and $F=(\frac p2,0)$ in part 1, and used those to express the distance between the circles' centers using the distance formula. The distance between the centers is also the sum of the given radii. Equating those two expressions of the distance yields $p^4+6p^3-8p^2-512=0$. And that's the sticking point. What should one do next? Or how else should one solve part 2?

I did solve it as follows, but am pretty sure this is not the intended solution:

That last equation yields $$(p^2+6p+8)p^2=16(p^2+32)$$Now, $p=4$ works by inspection. If $0<p\lessgtr4$ then $(p^2+6p+8)p^2\lessgtr(p^2+6\cdot4+8)16=16(p^2+32)$.

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A parabola could be defined in the following way:

parabola is a set of points, such that for any point $P$ of the set, the distance $|PF|$ to a fixed point $F$, the focus, is equal to the distance $P\ell$ to a fixed line $\ell$, the directrix

In part one of the question we were asked to find the distance between a point on the parabola $D$, and the directrix. Because you found that the coordinates of $D$ are $(\frac{32}{p},8)$, and we know that the directrix of a canonical parabola of this form is the line $x=-\frac{p}{2}$, we can conclude that the distance between $D$ and the directrix is $\frac{p}{2}+\frac{32}{p}$.

From part two, as you mentioned, we can infer that the distance between $D$ and the focus $F$ is the sum of the radii of the circles, which is $p+4+\frac{p}{2}$. But from the definition of a parabola those two lengths are equal, so we can conclude that $$p+4+\frac{p}{2}=\frac{p}{2}+\frac{32}{p}\ \Rightarrow\ (p+8)(p-4)=p^2+4p-32=0$$
It is given that $p>0$, so the only concise solution is $p=4$ as you got as well.

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