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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 a point of intersection of the circle and the parabola is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The focus of the parabola \mathrm{y^2-2 p x } is at \mathrm{\left(\frac{p}{2}, 0\right) }and its directrix is \mathrm{x=-\frac{p}{2} }. Since the circle touches the directrix, therefore.
Radius = Distance between the point \mathrm{ \left(\frac{p}{2}, 0\right) } and the line \mathrm{ x=-p / 2=p. }
So, the equation of the circle is \mathrm{ \left(x-\frac{p}{2}\right)^2+(y-0)^2=p^2 \quad x^2+y^2-p x-\frac{3 p^2}{4}=0.}
Solving this equation with \mathrm{ y^2=2 p x}, we obtain 

\begin{aligned} &\mathrm{x^2+p x-\frac{3 p^2}{4}=0}\\ &\mathrm{ \Rightarrow \quad(2 x-p)(2 x+3 p)=0 }\\ &\mathrm{ \Rightarrow \quad x=\frac{p}{2}, x=-\frac{3 p}{2} \quad \text { (at } \quad x=\frac{-3 p}{2} \text { we do not get real value of } y \text { ) }} \end{aligned}

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