Let P be the point on the parabola y 2 = 4 x which is at the shortest distance from the center S of the…

1. $SP=2\sqrt{5}$
2. $SQ :QP=\left(\sqrt{5}+1\right) :2$
3. The x - intercept of the normal to the parabola at P is 6
4. The slope of the tangent to the circle at Q is $\frac{1}{2}$

Solution

$y^2=4x$
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 $=\frac{1}{2}$

Asked in: JEE Advanced 2016 (Paper 2)

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