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 $\displaystyle x^2 = 4y$, 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!

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

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

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

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

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

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

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

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

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

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

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

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

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

The locus of $\displaystyle T$ is a horizontal line!

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

The midpoint has coordinates:

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

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

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

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