Question

Let $P$ be the point on the parabola $y^2=4x$, which is at the shortest distance from the centre 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)

The correct options are
A $SP=2\sqrt{5}$

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 $\frac{1}{2}$


Tangent to $y^2=4x$ at $(t^2,2t)$ is
$yt=x+t^2$
Equation of normal at P(t², 2t) is
$y+tx=2t+t^3$
Since, normal at P passes through centre of circle $S(2,8)$.
$\therefore 8+2t=2t+t^3$
$\Rightarrow t=2$, i.e. $P(4,4)$
This is because shortest distance between two curves lie along their common normal and the common normal will pass through the centre of circle.
$\therefore SP=\sqrt{(4-2{)}^2+(4-8{)}^2}=2\sqrt{5}$
$\therefore$ Option (a) is correct.
Also, $SQ=2$
$\therefore PQ=SP-SQ=2\sqrt{5}-2$
Thus, $\frac{SQ}{QP}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$
$\therefore$ Option (b) is wrong.
Now, $x$-intercept of normal is $x=2+2^2=6$
$\therefore$ Option (c) is correct.
Slope of tangent $=\frac{1}{t}=\frac{1}{2}$
$\therefore$ Option (d) is correct.