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Consider a circle with its centre lying on the focus of the parabola $\mathrm {y^2=2 p x}$ such that it touches the directrix of the parabola. Then the points of intersection of the circle and parabola are

Option: 1
$\mathrm {\left(\frac{p}{2}, p\right),\left(\frac{p}{2},-p\right)}$

Option: 2
$\mathrm {\left(\frac{p}{2},-\frac{p}{2}\right),\left(\frac{-p}{2}, p\right)}$

Option: 3
$\mathrm {\left(-\frac{p}{2}, p\right),\left(p, \frac{-p}{2}\right)}$

Option: 4
$\mathrm {\left(-\frac{p}{2},-\frac{p}{2}\right),\left(\frac{-p}{2}, \frac{p}{2}\right)}$

Answers (1)

best_answer

The focus of parabola $\mathrm {y^2=2 p x }$ is $\mathrm {\left(\frac{p}{2}, 0\right) }$ and directrix $\mathrm {x=-p / 2 }$

$\therefore$ Centre of circle is $\mathrm {\\ \left(\frac{p}{2}, 0\right) }$ and radius $\mathrm {\frac{p}{2}+\frac{p}{2}=p\\ }$

$\therefore$ Equation of circle is $\mathrm {\left(x-\frac{p}{2}\right)^2+y^2=p^2}$
For points of intersection with $\mathrm {y^2=2 p x}$ ........(1)

and $\mathrm {4 x^2+4 y^2-4 p x-3 p^2=0}$ ...........(2)

Can be obtained by solving (1) and (2) as follows

$\begin{aligned} & \mathrm {\Rightarrow 4 x^2+8 p x-4 p x-3 p^2=0 \Rightarrow 4 x^2+4 p x-3 p^2=0 }\\ & \mathrm {\Rightarrow \quad(2 x+3 p)(2 x-p)=0 \Rightarrow x=\frac{-3 p}{2}, \frac{p}{2} }\\ & \mathrm {\Rightarrow y^2=-3 p^2 \text { (not possible), } p^2 \Rightarrow y= \pm p }\\ \end{aligned}$
$\therefore$ Required points are $\mathrm {(p / 2, p),(p / 2,-p)}$

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