Question
Let $P(p^2, 2p)$, where $p > 0$, be a variable point on the curve whose equation is $y^2 = 4x$. The normal at P meets the curve again at the point $Q(q^2, 2q)$.
(i) Express $q$ in terms of $p$.
(ii) Prove that the length, $L$, of the line segment $PQ$ is given by $$L^2 = \frac {k(1 + p^2)^3} {p^4}\ ,$$ where $k$ is a constant to be determined.
My answer
Since $P$ is a variable point, it could be any point right? In particular, it can be $Q$, so $q = p$. However, this is clearly wrong, because I am unable to proceed and prove (ii).
Any hints/suggestions as to why my answer to (i) is wrong and what the correct answer is would be greatly appreciated :)
Edit
Following some help, I know that $q = -\sqrt {p^2 + 4}$.
\begin{align} L^2 & = (2p - 2q)^2 + (p^2 - q^2)^2 \\[5 mm] & = 4p^2 - 8pq + 4q^2 + 16 \\[5 mm] & = 8p^2 + 32 - 8pq \\[5 mm] & = 8p^2 + 32 + 8p\sqrt {p^2 + 4} \end{align}
I am not sure how to proceed from here to prove (ii). In particular, I notice that the required form does not have a square root, so how should I get rid of the square root in my answer?
First, let us find the implicit derivative of the curve $y^{2} = 4x$. Differentiating both sides, we have:
$$2yy' = 4$$
$$y' = \frac{2}{y}$$
Thus, the slope of the tangent line at the point $(p^{2}, 2p)$ is $\frac{1}{p}$, which means the normal is $-p$. Then, using point-slope form to find the equation of this normal line:
$$y - 2p = -p(x - p^{2})$$
$$y = -px + p^{3} + 2p$$
Now, we can substitute into the given equation for the curve to find the points of intersection:
$$(-px +p^{3} + 2p)^{2} = 4x$$
$$p^{2}x^{2} - (2p^{4} + 4p^{2} + 4)x +(p^{3} + 2p)^{2}= 0$$
Dividing both sides by $p^{2}$:
$$x^{2} - \frac{2p^{4} + 4p^{2} + 4}{p^{2}}x + (p^{2} + 2)^{2}$$
Because the normal intersects the curve a $(p^{2}, 2p)$, we know that $p^{2}$ must be a solution. Using this and the fact that the product of the roots is $(p^{2} + 2)^{2}$, the other solution is $x = \big(\frac{p^{2} + 2}{p}\big)^{2}$.
Now, since it is given that $p > 0$, we note that the $y$-coordinate of the second intersection point must be negative. (Think of this as a line intersecting both branches of a sideways parabola, one branch above the $x$-axis and one part below). Then, we must have:
$$\boxed{(q^{2}, 2q) = \bigg(\bigg(\frac{p^{2} + 2}{p}\bigg)^{2}, -\frac{2p^{2} + 4}{p}\bigg)\text{ and }q = -\frac{2p^{2} + 4}{p}}$$
Now, using the distance formula:
$$PQ = L = \sqrt{\bigg(p^{2} - \bigg(\frac{p^{2} + 2}{p}\bigg)^{2}\bigg)^{2} + \bigg(2p + \frac{2p^{2} + 4}{p}\bigg)^{2}}$$
$$L^{2} = \bigg(p^{2} - \bigg(\frac{p^{2} + 2}{p}\bigg)^{2}\bigg)^{2} + \bigg(2p + \frac{2p^{2} + 4}{p}\bigg)^{2}$$
Multiplying out by $p^{4}$ to clear denominators:
$$p^{4}L^{2} = (p^{4} - (p^{2} + 2)^{2})^{2} + (2p^{3} + 2p^{3} + 4p)^{2}$$
$$p^{4}L^{2} = (-4p^{2} - 4)^{2} + (4p^{3} + 4p)^{2}$$
$$p^{4}L^{2} = (-4)^{2}(p^{2} + 1)^{2} + (4p)^{2}(p^{2} + 1)^{2}$$
$$p^{4}L^{2} = (16 + 16p^{2})(p^{2} + 1)^{2}$$
$$p^{4}L^{2} = 16(p^{2}+ 1)^{3}$$
$$\boxed{L^{2} = \frac{16(p^{2} + 1)^{3}}{p^{4}}, k = 16}$$