P is a point on the parabola whose ordinate equals its abscissa. A...

Let, $\text{P}\left(a{t}^{2}, 2at\right) be a point on$ the parabola{y}^{2} = 4ax$<br />Since, the ordinate equals its abscissa.<br />$⇒ a{t}^{2} = 2at ⇒ t = 2$<br />$P ≡ \left(4a, 4a\right)$<br />Equation of the normal at$\text{P}\left(4a, 4a\right) is
$y + 2x = 2a \left(2\right)$ + a{\left(2\right)}^{3}
$⇒ y + 2x = 12a \ldots$\left(i\right)$<br />${y}^{2} = 4ax$<br />$⇒ {y}^{2} = 2a\left(12a - y\right) … [From \left(i\right)]
$⇒ {y}^{2} + 2ay - 24{a}^{2} = 0$
$⇒ \left(y - 4a\right) \left(y + 6a\right)$ = 0$<br />$⇒ y = 4a or y = - 6a$<br />$⇒ $\text{S}$ ≡ \left(a, 0\right)$,$\text{P}$≡ \left(4a, 4a\right)$, $\text{Q}$ ≡ \left(9a, - 6a\right)$<br />Slope of$\text{SP} =\dfrac{4}{3} and slope of\text{SQ} =\dfrac{- 6}{8}$<br />$⇒$Required product$= \dfrac{4}{3} \times \dfrac{- 6}{8} =- 1