The point $ P \left( \frac{2}{p}, \frac{1}{p^2} \right) $, where $ p \neq 0 $, lies on the parabola $ x^2 = 4y $ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2017 - Paper 1

Substituting $x = \frac{2}{p}$ into the parabola equation:

$\left( \frac{2}{p} \right)^2 = 4 \left( \frac{1}{p^2} \right)$

Calculating:

$\frac{4}{p^2} = \frac{4}{p^2}$

This is true, confirming that point $P$ lies on the parabola.

To find the slope of the tangent at point $P$, we differentiate the parabola equation:

$x^2 = 4y \implies 2x \frac{dx}{dt} = 4 \frac{dy}{dt} \implies \frac{dy}{dx} = \frac{x}{2}$

At $P \left( \frac{2}{p}, \frac{1}{p^2} \right)$, the slope is:

$\frac{dy}{dx} = \frac{\frac{2}{p}}{2} = \frac{1}{p}$

The slope of the normal line is the negative reciprocal of the tangent slope:

$m_{normal} = -p$

Using point-slope form of the line equation:

$y - y_1 = m (x - x_1)$

Where $(x_1, y_1) = \left( \frac{2}{p}, \frac{1}{p^2} \right)$ and $m = -p$, we have:

$y - \frac{1}{p^2} = -p \left( x - \frac{2}{p} \right)$

Expanding and rearranging gives us the equation of the normal:

$p^2y + px = 1 + 2p^2$