Question

Consider a circle with its centre lying on the focus of the parabola $y^2=2px$ such that it touches the directrix of the parabola. Then the point of intersection of the circle and parabola can be

• A. $(-\frac{p}{2},p)$
• B. $(\frac{p}{2},p)$
• C. $(\frac{p}{2},-p)$
• D. $(\frac{p}{2},-\frac{p}{2})$

Answer 1

Answer: (B), (C)

$(\frac{p}{2},-p)$

The focus of the parabola $y^2=2px$ is $(\frac{p}{2},0)$ and directrix is $x=-\frac{p}{2}$.
$\therefore$ Centre of circle is $(\frac{p}{2},0)$
and radius $=\frac{p}{2}+\frac{p}{2}=p$
$\therefore$Equation of circle is ${(x-\frac{p}{2})}^2+y^2=p^2$
Solving with $y^2=2px,$ we get
${(x-\frac{p}{2})}^2+2px-p^2=0$
$\Rightarrow 4x^2+8px-4px-3p^2=0$
$\Rightarrow 4x^2+4px-3p^2=0$
$\Rightarrow 4x^2+6px-2px-3p^2=0$
$\Rightarrow (2x-p)(2x+3p)=0$
$\Rightarrow x=\frac{p}{2}$ or $-\frac{3p}{2}$

For $x=\frac{p}{2},$
$y^2=2p\times \frac{p}{2}$
$\Rightarrow y=\pm p$

For $x=-\frac{3p}{2},$
$y^2=2p\times \frac{-3p}{2}=-6p^2$ (not possible)
$\therefore$ Required points are $(\frac{p}{2},\pm p)$