1. ## Parametric Equations

I'm having problems with a few parametric equation questions. Considering how much trouble I've been having with this particular unit of work, I expect I'll have to use this thread to ask more questions in the near future. The one I'm stuck on at the moment is;

PQ, a variable chord of the parabola $x^2 = 4y$, subtends a right angle at the vertex O.

a) If p and q are parameters corresponding to P and Q, show that pq = -4.

b) If the tangents at P and Q intersect at T, find the locus of T.

c) If M is the midpoint of chord PQ, determine the locus of M.

2. Hello, Flay!

$PQ$, a variable chord of the parabola $x^2 \,= \,4y$, subtends a right angle at the vertex $O.$

a) If $p\text{ and }q$ are parameters corresponding to $P$ and $Q$, show that: $pq = -4.$
Parametric equations: . $\begin{array}{ccc}x &=& 2t \\ y &=&t^2 \end{array}$

The points are: . $\begin{array}{c}P\,(2p,p^2) \\ Q\,(2q,q^2) \end{array}$

The slope of $OP$ is: . $m_{OP} \:=\:\frac{p^2 - 0}{2p-0} \:=\:\frac{p}{2}$

The slope of $OQ$ is: . $m_{OQ} \:=\:\frac{q^2-0}{2q-0} \:=\:\frac{q}{2}$

Since $OP \perp OQ$, we have: . $\frac{p}{2} \:=\:-\frac{2}{q}\quad\Rightarrow\quad\boxed{ pq \:=\:-4}$

Note: .Since $q = -\frac{4}{p}$, point $Q$ becomes: . $\left(-\frac{8}{p},\:\frac{16}{p^2}\right)$

b) If the tangents at $P\text{ and }Q$ intersect at $T$, find the locus of $T.$
The slope of the tangent is: . $\frac{dy}{dx} \:=\:\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \:=\:\frac{2t}{2} \:=\:t$

At $P$, we have the point $(2p, p^2)$ and slope $p.$
The equation of the tangent at $P$ is:
. . $y - p^2 \:=\:p(x-2p)\quad\Rightarrow\quad y \:=\:px - p^2\;\;{\color{blue}[1]}$

At $Q$, we have the point $\left(\text{-}\frac{8}{p},\:\frac{16}{p^2}\right)$ and slope $-\frac{4}{p}$
The equation of the tangent at Q is:
. . $y - \frac{16}{p^2} \:=\:-\frac{4}{p}\left(x + \frac{8}{p}\right)\quad\Rightarrow\quad y \:=\:-\frac{4}{p}x - \frac{16}{p^2}\;\;{\color{blue}[2]}$

The two tangents intersect: . $px-p^2 \:=\:-\frac{4}{p}x - \frac{16}{p^2} \quad\Rightarrow\quad px + \frac{4}{p}x \:=\:p^2-\frac{16}{p^2}$

. . Factor: . $\left(p + \frac{4}{p}\right)x \:=\:\left(p - \frac{4}{p}\right)\left(p + \frac{4}{p}\right) \quad\Rightarrow\quad x \:=\:p - \frac{4}{p}$

Substitute into [1]: . $y \:=\:p\left(p - \frac{4}{p}\right) - p^2 \quad\Rightarrow\quad \boxed{y \:=\:-4}$

The locus of $T$ is a horizontal line!

c) If $M$ is the midpoint of chord $PQ$, determine the locus of $M.$
We have: . $P(2p,p^2)\,\text{ and }\,Q\left(-\frac{8}{p},\:\frac{16}{p^2}\right)$

The midpoint has coordinates:

. . $x \;=\;\frac{2p-\frac{8}{p}}{2} \;=\;p - \frac{4}{p}\;\;{\color{blue}[3]}\qquad\qquad y \;=\;\frac{p^2 + \frac{16}{p^2}}{2} \;=\;\frac{p^4+16}{2p^2}\;\;{\color{blue}[4]}$

Square [3]: . $x^2 \;\;=\;\;p^2 - 8 + \frac{16}{p^2} \;\;=\;\;\frac{p^4 + 16}{p^2} - 8 \;\;=\;\;2\left(\frac{p^4+16}{2p^2}\right) - 8$

Substitute [4]: . $\boxed{x^2 \;=\;2y - 8}$

The locus of $M$ is another parabola,
. . somewhat "narrrower" with vertex (0,4).