Question

Consider a circle with center lying on the focus of the parabola $y^2=2px$ such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is

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

Answer 1

Answer: (A), (B)

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

Focus of the parabola $y^2=2px$ is $(\frac{p}{2},0)$
$\therefore$ center of the circle is $(\frac{p}{2},0)$
Radius of the circle is ${(x-\frac{p}{2})}^2+{(y-0)}^2=p^2$ ............$\left(1\right)$
Solving eqn$\left(1\right)$ and $y^2=2px$ then
${(x-\frac{p}{2})}^2+2px=p^2$
On expanding, we get
$\Rightarrow x^2+\frac{p^2}{4}-px+2px=p^2$
$\Rightarrow x^2+px=\frac{3p^2}{4}$
$\Rightarrow 4x^2+4px=3p^2$
On factorising , we get
$\Rightarrow 4x^2+4px-3p^2=0$
$\Rightarrow (2x+3p)(2x-p)=0$
$\Rightarrow 2x=p,2x=-3p$ or $x=\frac{p}{2},\frac{-3p}{2}$
$\therefore$ when $x=\frac{p}{2}$ we have $y^2=2px=2p\left(\frac{p}{2}\right)=p^2$
When $x=\frac{-3p}{2}$ we have $y^2=2px=2p\left(\frac{-3p}{2}\right)$
$\Rightarrow y=\sqrt{{-3p}^2}$ is imaginary which is impossible.
$\therefore y^2=p^2$
$\Rightarrow y=\pm p$
$\therefore$ points of intersection are $(\frac{p}{2},p)$ and $(\frac{p}{2},-p)$