Question

$P$ is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at $P$ to meet it again at $Q.$ If $S$ is the focus of the parabola, then the product of the slopes of $SP$ and $SQ$ is

Answer 1

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