Question

$P$ is a point on the parabola $y^2=4ax$ 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 absolute value of the product of the slopes of $SP$ and $SQ$ is

Answer 1

Answer: 1

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