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)

$SP=2\sqrt{5}$
The $x$-intercept of the normal to the parabola at $P$ is $6$
The slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Let any point $P(t^2,2t)$ on parabola.
As we know shortest distance between two curves lies along their common normal.
The common normal will pass through centre of circle.
Slope of normal to the parabola $y^2=4x$ at $P=t,\frac{2t-8}{t^2-2}=-t$
$\Rightarrow t^3=8\Rightarrow t=2\therefore P(4,4)$

$\therefore$ $SP=\sqrt{(4-2{)}^2+(4-8{)}^2}=2\sqrt{5}$
(i) equation of normal at $P(4,4)$ $\Rightarrow y=2x+12\Rightarrow$ x-intercept $=6$
(ii) slope of tangent at $Q=$ slope of tangent at $P$ $=\frac{1}{2}$
(iii) $\frac{SQ}{PQ}=\frac{2}{2\sqrt{5}-2}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$