Question

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then

• A. $SP=2\sqrt{5}$
• B. $SQ:QP=(\sqrt{5}+1):2$
• C. the $x$-intercept of the normal to the parabola at $P$ is $6$
• D. the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Answer 1

Answer: (A), (C), (D)

$P\left(t^2,2t\right)$ centre of circle $S(2,8)$
Distance $SP=\sqrt{{\left(t^2-2\right)}^2+(2t-8{)}^2}$
$z={\left(t^2-2\right)}^2+(2t-8{)}^2$
$\frac{dz}{dt}=2\left(t^2-2\right)\cdot (2t)+2(2t-8)\cdot 2$
for minima $\Rightarrow \frac{dz}{dt}=0$
$\Rightarrow 4\left[t^3-2t+2t-8\right]=0$
$\Rightarrow t=2$
$\therefore P(4,4);S(2,8)$
(A) $SP=2\sqrt{5}$
(B) $\frac{SQ}{QP}=\frac{2}{2\sqrt{5}-2}=\frac{1}{\sqrt{5}-1}$
$\frac{SQ}{QP}=\frac{\sqrt{5}+1}{4}$
(C) Parabola $y^2=4x$
$\Rightarrow 2y\frac{dy}{dx}=4$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(4,4)}=\frac{1}{2}$
Equation of normal $y-4=-2(x-4)$
$\therefore x$ intercept $=6$
(D) Slope of $SP=\frac{8-4}{2-4}=-2$
$\therefore$ Slope of tangent at $Q$ is $\frac{1}{2}$
$(\because$ tangent is perpendicular to SP).