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If a circle with its centre at the focus of the parabola $y^2=2 p x$ is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is
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The correct answer is:
$\left(\frac{p}{2},-p\right)$
Equation of given parabola is $y^2=2 p x$ having focus $F\left(\frac{p}{2}, 0\right)$ and equation of directrix is
$x+\frac{p}{2}=0$, so equation of circle having centre $\left(\frac{p}{2}, 0\right)$ and radius $r=p$, because the circle touches the directrix of the given parabola, is
$(x-p / 2)^2+y^2=p^2$ $\ldots(i)$
Now, on solving the equation of the given parabola and circle (i), we get
$\left(x^2-p x+\frac{p^2}{4}\right)+2 p x=p^2 \Rightarrow\left(x^2+p x+\frac{p^2}{4}\right)=p^2$
$\Rightarrow \quad\left(x+\frac{p}{2}\right)^2=p^2 \Rightarrow x+\frac{p}{2}= \pm p$
$\Rightarrow \quad x=\frac{p}{2} \quad[\because x p>0]$
and $\quad y= \pm p$
$\therefore$ Point of intersections of circle and parabola are $\left(\frac{p}{2}, p\right),\left(\frac{p}{2},-p\right)$
$x+\frac{p}{2}=0$, so equation of circle having centre $\left(\frac{p}{2}, 0\right)$ and radius $r=p$, because the circle touches the directrix of the given parabola, is
$(x-p / 2)^2+y^2=p^2$ $\ldots(i)$
Now, on solving the equation of the given parabola and circle (i), we get
$\left(x^2-p x+\frac{p^2}{4}\right)+2 p x=p^2 \Rightarrow\left(x^2+p x+\frac{p^2}{4}\right)=p^2$
$\Rightarrow \quad\left(x+\frac{p}{2}\right)^2=p^2 \Rightarrow x+\frac{p}{2}= \pm p$
$\Rightarrow \quad x=\frac{p}{2} \quad[\because x p>0]$
and $\quad y= \pm p$
$\therefore$ Point of intersections of circle and parabola are $\left(\frac{p}{2}, p\right),\left(\frac{p}{2},-p\right)$
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