# Thread: Parametric Equations of the Parabola

1. ## Parametric Equations of the Parabola

'm having trouble with part of a particular parametric parabola question. Please remember that this is not the full question, only part;

Prove that SP = a(p^2 + 1) units [where S = (0, a) or the focus, and P = (2ap, ap^2)] and hence that SZ^2 = OS * SP [where O is the origin].

2. Prove $SP = a(p^2+1)$

Using the distance formula, with points $S(0,a)$ and $P(2ap,ap^2)$

$SP^2 = (0-2ap)^2 + (a-ap^2)^2$
= $4a^2p^2 + a^2 (1-p^2)^2$
Factorising and expanding, we get
= $a^2 (4p^2 + 1 - 2p^2 + p^4)$
= $a^2 (p^4 + 2p^2 + 1)$
$SP^2$= $a^2 (p^2+1)^2$
$SP = a(p^2+1)$

For the second one, would you mind telling us what the point Z is?

3. Thanks for that, I'm alright for the rest of that particular part. The problem is, I'm having more troubles with the later parts of this question. Maybe it'd be better if I typed out the entire thing...

The equation of the tangent at P(2ap, ap^2) on the parabola x^2 = 4ay (whose vertex is 0) is y - px + ap^2 = 0. The tangent meets the x-axis at Z (ap, 0) and the y-axis at T (0, -ap^2). PN is drawn perpendicular to the axis of the parabola meeting it at N (0, ap^2), and PM meets the directrix at right angles in M.

iv) Prove that the tangent PT makes equal angles wih the axis of the parabola and wih PS (HINT: show triangle PST is isosceles)

v) Prove the lines SM, TP intersect at right angles at Z.

4. iv) Prove that the tangent PT makes equal angles wih the axis of the parabola and wih PS (HINT: show triangle PST is isosceles)
Using the hint given, lets try to prove triangle PST is isoceles. Easiest way to prove this is prove SP = TS.

From above, we know that SP = $a(p^2+1)$

$ST^2 = (0-0)^2 + (a-(-ap^2))^2$
$= (a(1+p^2))^2$
$ST = a(p^2+1)$

Therefore ST = SP; and triangle PST is isoceles (1 pair of equal sides).
Thus, tangent PT makes equal angles with axis of parabola with PS (base angles of triangle PST).

v) Prove the lines SM, TP intersect at right angles at Z.
Using gradient formula, gradient of line TP (where T = (0, apē); P = (2ap, apē) )is
$M_{TP} = \frac{ap^2-(-ap^2)}{2a-0}$
= $\frac{2ap^2}{2ap}$
= $p$

Point M has directly below point P for PM to intersect the directrix, which means it has the same x value (2ap). M also lies on the directrix, which has the y value (-a)
Gradient of line SM (where S = (0,a) M = (2ap, -a))is:
$M_{SM} = \frac{a-(-a)}{0-2ap}$
$= \frac{2a}{-2ap}$
= $\frac{1}{-p}$

Gradient SM = $\frac{1}{-p}$ and Gradient TP = $p$