Question

From the focus of the parabola $y^2=2px$ as centre a circle is described so that a common chord of the curves is equidistant from the vertex and the focus of the parabola. Find the equation of the circle.

• A. ${(x-\frac{p}{2})}^2+y^2=\frac{9p^2}{16}$
• B. ${(x-\frac{p}{4})}^2+y^2=\frac{9p^2}{16}$
• C. ${(x-\frac{p}{2})}^2+y^2=\frac{6p^2}{16}$
• D. ${(x-\frac{p}{2})}^2+y^2=\frac{8p^2}{16}$

Answer 1

The correct option is A ${(x-\frac{p}{2})}^2+y^2=\frac{9p^2}{16}$

Given parabola is $y^2=2px$, thus focus is $S(p/2,0).$
Circle with centre $(p/2,0)$ is
${(x-p/2)}^2+y^2=r^2$
We have to find the radius
Let AB is the common chord, thus $(p/4,0)$ is mid - point
$\Rightarrow LS=p/4=OL=x$(say)
$\Rightarrow A{L}^2=y^2$ where $y^2=2p.\frac{p}{4}=\frac{p^2}{2}$
$\therefore r^2=A{L}^2+S{L}^2=\frac{p^2}{2}+\frac{p^2}{16}=\frac{9p^2}{16}\therefore r=\frac{3p}{4}$
Hence the circle is ${(x-\frac{p}{2})}^2+y^2=\frac{9p^2}{16}$