If the normal at $P \left(18, 12\right)$ to the parabola $y^2 = 8x$ cuts it again at $Q$, show that $\;$ $9PQ = 80 \sqrt{10}$
Equation of parabola: $\;$ $y^2 = 8x$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives
$4a = 8$ $\implies$ $a = 2$
Equation of normal to the parabola $\;$ $y^2 = 4ax$ $\;$ at $\;$ $P \left(x_1, y_1\right)$ $\;$ is
$y - y_1 = \left(\dfrac{-y_1}{2a}\right) \left(x - x_1\right)$
Given point: $\;$ $P \left(18, 12\right)$
$\therefore \;$ Equation of normal at $P \left(18, 12\right)$ to parabola $(1)$ is
$y - 12 = \left(\dfrac{-12}{2 \times 2}\right) \left(x - 18\right)$
i.e. $\;$ $y - 12 = -3x + 54$
i.e. $\;$ $3x + y - 66 = 0$ $\;\;\; \cdots \; (2)$
To determine the point $Q$ where the normal again meets the parabola, substitute $\;$ $y = 66 - 3x$ $\;$ from equation $(2)$ in equation $(1)$
$\therefore \;$ We have,
$\left(66 - 3x\right)^2 = 8x$
i.e. $\;$ $4356 + 9x^2 - 396x = 8x$
i.e. $\;$ $9x^2 - 404x + 4356 = 0$
Solving, we get, $\;$ $x = \dfrac{242}{9}$ $\;$ or $\;$ $x = 18$
When $\;$ $x = \dfrac{242}{9}$, $\;$ we have from equation $(2)$
$y = 66 - 3 \times \dfrac{242}{9} = \dfrac{-44}{3}$
$\therefore \;$ $Q = \left(\dfrac{242}{9}, \dfrac{-44}{3}\right)$
$\therefore \;$ $PQ = \sqrt{\left(18 - \dfrac{242}{9}\right)^2 + \left(12 + \dfrac{44}{3}\right)^2}$
i.e. $\;$ $PQ = \sqrt{\dfrac{6400}{81} + \dfrac{6400}{9}} = 80 \sqrt{\dfrac{1 + 9}{81}} = \dfrac{80 \sqrt{10}}{9}$
$\implies$ $9 PQ = 80 \sqrt{10}$
Hence proved.