# The normal (2ap, ap^2) to the parabola x^2=4ay meets the curve again at Q(2aq, aq^2)?

## (a) Show that $q = - \frac{2 + {p}^{2}}{p}$ (b) Find the coordinates of P so that the lines $O P$ and $O Q$ meet at right angles, where $O$ is the origin.

Nov 20, 2017

(a)Given equation of the parabola
${x}^{2} = 4 a y$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4 a} \cdot 2 x = \frac{x}{2 a}$

Slope of the normal

$m = - \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}} _ \left(2 a p , a {p}^{2}\right) = - \frac{2 a}{2 a p} = - \frac{1}{p}$

Again the slope of the normal as it joins $\left(2 a p , a {p}^{2}\right) \mathmr{and} \left(2 a q , a {q}^{2}\right)$ is also
$m = \frac{a {p}^{2} - a {q}^{2}}{2 a p - 2 a q} = \frac{1}{2} \left(p + q\right)$

So we have

$\frac{1}{2} \left(p + q\right) = - \frac{1}{p}$

$\implies q = - \frac{2}{p} - p = - \frac{2 + {p}^{2}}{p} \ldots . \left(1\right)$

OP and OQ being orthogonal the produt of their slopes $\frac{a {p}^{2}}{2 a p} \cdot \frac{a {q}^{2}}{2 a q} = - 1$

$\implies p q = - 4. \ldots \left(2\right)$

From (1) and (2) we get

$- 4 = - \left(2 + {p}^{2}\right)$

$\implies p = \sqrt{2}$

(b)So coordinates of the point P will be $\left(2 \sqrt{2} a , 2 a\right)$