Question

Consider the parabola $x^2=4y$ 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 $△PQR$ is $\left(0,\frac{7}{9}\right)$.
• B. Distance between chord $PQ$ and directrix of the parabola is $\frac{4}{3}$.
• C. Area of $△PQR$ 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 $x^2=4y$ in equation of given circle, we get $4y+y^2-10y+\left(25-r^2\right)=0$
$\Rightarrow y^2-6y+\left(25-r^2\right)=0$
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 $y=mx+c$ be tangent to $x^2=4y$, so on solving we get
$x^2-4mx-4c=0$
Put discriminant to zero, we get $m^2=-c$ (condition of tangency) $\therefore y=mx-m^2$ is tangent to parabola $x^2=4y$.
But by applying condition of tangency of above tangent with the given circle, we get
$\frac{|5+m^2|}{\sqrt{1+m^2}}=4\Rightarrow m^4-6m^2-9=0\Rightarrow {\left(m^2-3\right)}^2=0\Rightarrow m=\pm \sqrt{3}$
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 $SM=5-\frac{1}{3}=\frac{14}{3}$.
Clearly centroid of $△PQR$ is $\left(0,\frac{-7}{6}\right)$ and area of $△PQR$ is $\frac{20}{3\sqrt{3}}$.