Question

Let $P$ be the point on the parabola $y^{2}=4 x$ which is at the shortest distance from the center $S$ of the circle $x ^{2}+ y ^{2}-4 x -16 y +64=0$. 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)

$P \left( t ^{2}, 2 t \right) \quad$ centre of circle $S (2,8)$
Distance $SP =\sqrt{\left( t ^{2}-2\right)^{2}+(2 t -8)^{2}}$
$z=\left(t^{2}-2\right)^{2}+(2 t-8)^{2} $
$\frac{d z}{d t}=2\left(t^{2}-2\right) \cdot(2 t)+2(2 t-8) \cdot 2$
for minima $\Rightarrow \frac{ d z}{ dt }=0$
$\Rightarrow 4\left[t^{3}-2 t+2 t-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 $y^{2}=4 x$
$\Rightarrow 2 y \frac{d y}{d x}=4 $
$\Rightarrow \left(\frac{d y}{d x}\right)_{(4,4)}=\frac{1}{2}$
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).