Mathematics - Parabola Question with Solution | TestHub

$y^2=4x$
Point P lies on normal to parabola passing through centre of circle
$y+tx=2t+t^3$......(i)
$8+2t=2t+t^3$
$t=2$
$P\left(4, 4\right)$
$SP= \sqrt{{\left(4-2\right)}^2+{\left(4-8\right)}^2}$
$SP=2\sqrt{5}$
$SQ=2$
$\Rightarrow$$PQ=2\sqrt{5}-2$
$\frac{SQ}{QP}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$
To find$x$intercept
Put$y=0$in (i)
$\Rightarrow$$x=2+t^2$
$x=6$
$\because$Slope of common normal$= -t= -2$
$\therefore$Slope of tangent$=\frac{1}{2}$