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  • Consider a circle with its centre lying on the focus of the parabola such that it touches t

Consider a circle with its centre lying on the focus of the parabola y^2=2 p x such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is

Option: 1

\left(\frac{p}{2}, p\right)


Option: 2

\left(\frac{p}{2},-p\right)


Option: 3

\left(-\frac{p}{2}, p\right)


Option: 4

\left(-\frac{p}{2},-p\right)


Answers (1)

best_answer

Focus of the parabola y^2=2 p x \text { is }\left(\frac{p}{2}, 0\right)

\therefore Centre of circle is \left ( \frac{p}{2},0 \right )

Radius of the circle = Distance between focus and directrix                                   = semi latusrectum
                                =p
\therefore Equation of circle is

                    \left(x-\frac{p}{2}\right)^2+(y-0)^2=p^2\: \: \: \: \: \: \: \: \: ...(i)

Solving Eq. (i) and y^2=2 p x \text {, then }

                         \begin{array}{r} \left(x-\frac{p}{2}\right)^2+2 p x=p^2 \\ \\\Rightarrow x^2+p x=\frac{3 p^2}{4} \\ \\\Rightarrow 4 x^2+4 p x-3 p^2=0 \end{array}

  \begin{array}{lll} \therefore & x=p / 2 \text { and } x=-3 p / 2 \\ \\\because & y^2=2 p x=2 p\left(\frac{p}{2}\right) \end{array}

and y^2=2 p\left(\frac{-3 p}{2}\right)=-3 p^2, y is imaginary (Impossible) 

\therefore \: \: \: \: \: \: \: \: \mathrm{y^{2}=P^{2}}

\Rightarrow \: \: \: \: \: \: \: \: \: \: \mathrm{y=\pm\: p}

\therefore Point of intersection are \left(\frac{p}{2}, p\right) \text { and }\left(\frac{p}{2},-p\right)

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