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 $\frac{1}{2}$

The centre of circle is $S(2,8)$

Let $P=(t^2,2t)$

For point P to be closest to S, SP must normal to parabola at P

Slope of normal to parabola at P is $-t$

Therefore, slope of SP $=-t$

$\therefore \frac{8-2t}{2-2t^2}=-t$

$\therefore t=2$

So, $P=(4,4)$

a) $S=(2,8)$, $P=(4,4)$

$\therefore SP=\sqrt{(4-2{)}^2+(4-8{)}^2}=2\sqrt{5}$

c) Slope of normal to parabola at P $=-t=-2$

So, the equation of normal is $(y-4)=-2(x-4)$

$\therefore y=12-2x$

$⟹x-intercept=6$

b) Q is the point of intersection of circle and the line $y=12-2x$

Solving the two equations, we get

$Q=(2+\frac{2}{\sqrt{5}},2+\frac{2}{\sqrt{5}})$

$SQ=\sqrt{{(2-(2+\frac{2}{\sqrt{5}}\left)\right)}^2+{(8-(2+\frac{2}{\sqrt{5}}\left)\right)}^2}$

$\therefore SQ=\sqrt{24(1+\frac{1}{\sqrt{5}})}$

$\therefore \frac{SQ}{PQ}=\frac{2}{\sqrt{24(1+\frac{1}{\sqrt{5}})}}$

d) Slope of SQ=slope of SP=-2

SQ is normal to the circle at Q

So, slope of tangent at Q$=\frac{-1}{-2}=\frac{1}{2}$

So, A,C,D are the right answers