Question

A circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum of a parabola; prove that the common chord of the circle and parabola bisects the distance between the vertex and the focus.

Answer 1

Let the parabola be $y^2=4ax$

Latus rectum$=4a$

Diameter of circle $=\frac{3}{4}\times 4a=3a$

Radius $=\frac{3a}{2}$

Equation of circle with centre as vertex is

$x^2+y^2=\frac{{9a}^2}{4}$

Equation of common chord is obtained by subtracting equation of both curves

$4x^2={9a}^2-16ax4x^2+16ax-{9a}^2=04x^2-2ax+18ax-{9a}^2=02x(2x-a)+9a(2x-a)=0(2x+9a)(2x-a)=0\Rightarrow x=\frac{a}{2},-\frac{9a}{2}$

You can clearly see from the figure $x=-\frac{9a}{2}$ is not possible as the curves do not meet at negative values of $x$

So the equation of common chord is

$x=\frac{a}{2}$

Mid point of vertex and focus is $(\frac{a+0}{2},\frac{0+0}{2})\Rightarrow (\frac{a}{2},0)$

Mid point lies on the common chord, so the chord bisect the distance between vertex and focus.