Question

Consider the parabola $x^2=4 y$ and circle $x^2+(y-5)^2=r^2(r>0)$. Given that the circle touches the parabola at the points $P$ and $Q$. Let $R$ be the point of intersection of tangents to parabola at $P$ and $Q$ and $S$ be the centre of circle.
Which of the following statement(s) is/are TRUE?

• A. Centroid of $\triangle PQR$ is $\left(0, \frac{7}{9}\right)$.
• B. Distance between chord $PQ$ and directrix of the parabola is $\frac{4}{3}$.
• C. Area of $\triangle PQR$ is $\frac{40}{3 \sqrt{3}}$.
• D. The common tangents to the circle and parabola intersect at $(0,-3)$.

Answer 1

Answer: (B), (D)

Putting $x ^2=4 y$ in equation of given circle, we get $4 y + y ^2-10 y +\left(25- r ^2\right)=0$
$\Rightarrow y^2-6 y+\left(25-r^2\right)=0$
As parabola is tangent to circle, so discriminant of above equation must be zero.

$\therefore 36=4\left(25-r^2\right) \Rightarrow r=4=$ radius of circle.
Let $y=m x+c$ be tangent to $x^2=4 y$, so on solving we get
$x^2-4 m x-4 c=0$
Put discriminant to zero, we get $m ^2=- c$ (condition of tangency) $\therefore y = mx - m ^2$ is tangent to parabola $x ^2=4 y$.
But by applying condition of tangency of above tangent with the given circle, we get
$\frac{\left|5+ m ^2\right|}{\sqrt{1+ m ^2}}=4 \Rightarrow m ^4-6 m ^2-9=0 \Rightarrow\left( m ^2-3\right)^2=0 \Rightarrow m = \pm \sqrt{3}$
Points of contact to the parabola will be $P \left(\frac{2}{ m _1}, \frac{1}{ m _1^2}\right)$ & $Q \left(\frac{2}{ m _2}, \frac{1}{ m _2^2}\right)$ i.e. $P \left(\frac{2}{\sqrt{3}}, \frac{1}{3}\right)$ & $Q \left(\frac{-2}{\sqrt{3}}, \frac{1}{3}\right)$.
$\therefore$ Equation of common tangents will be $y= \pm \sqrt{3} x-3$ which intersect at $R(0,-3)$
Center of circle is $S (0,5)$ & equation of chord $PQ$ is $y =\frac{1}{3}$.
Equations of latus-rectum and directrix of parabola will be $y=1$ and $y=-1$ respectively which are at distances $\frac{2}{3}$ and $\frac{4}{3}$ respectively. Also distance $S M=5-\frac{1}{3}=\frac{14}{3}$.
Clearly centroid of $\triangle PQR$ is $\left(0, \frac{-7}{6}\right)$ and area of $\triangle PQR$ is $\frac{20}{3 \sqrt{3}}$.