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Consider a circle with its center 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 parabola is 

Option: 1

\begin{aligned} & \left(\frac{p}{2}, p\right) \text { or }\left(\frac{p}{2},-p\right) \\ \end{aligned}


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The focus of the parabola y^2=2 p x is  (\frac{p}{2},0)

And directrix is .x=-\frac{p}{2}

Here, p>0

Center of the circle is  (\frac{p}{2},0)

and radius is .\frac{p}{2}+\frac{p}{2}=p

Equation of the circle is .\left(x-\frac{p}{2}\right)^2+y^2=p^2

For pts of intersection of  y^2=2 p x  —--(i)

And  4 x^2+4 y^2-4 p x-3 p^2=0—-------(ii)

From (i) and (ii)

\begin{aligned} & 4 x^2+8 p x- 4 p x-3 p^2=0 \\ \Rightarrow & (2 x+3 p)(2 x-p)=0 \\ \Rightarrow & x=\frac{-3 p}{2}, \frac{p}{2} \\ \Rightarrow & y=-3 p^2 \end{aligned}( Not Possible)

\begin{aligned} & \Rightarrow y=p^2 \\ & \Rightarrow y= \pm p \end{aligned}

So the required points are \left(\frac{p}{2}, p\right) \text { or }\left(\frac{p}{2},-p\right)

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Gunjita

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