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

Answer: -1

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