Question

A is the vertex and S is the focus of the parabola $y^2=4ax$. Prove that the common chord of the parabola and the circle, centre A and radius equal to 3/8th of the length of the latus rectum of the parabola is the perpendicular bisector of AS.

Answer 1

Given parabola: $y^2=4\times a\times x⟶\left(1\right)$

vertex, $A:(0,0)$

Focus, $S:(a,0)$

Circle with centre $A:(0,0)$ radius$=\frac{3}{8}$ (Length of L.R parabola)

radius$=\frac{3}{4}(4\times a)$

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

Equation of circle: $x^2+y^2=\frac{9}{4}{a}^2\to \left(2\right)$

Point of intersection of $\left(1\right)$ and $\left(2\right)$, $x^2+4ax=\frac{9}{4}{a}^2$

$x=\frac{1}{2}a$ or $x=-\frac{9}{2}a$ (not possible)

$y=\pm \sqrt{2a}$

Point of intersection: $(\frac{1}{2a},\sqrt{2a})$ & $(\frac{1}{2}a,-\sqrt{2a})$

Line passing through these two points will have equation $=(y-\sqrt{2a})=\frac{1}{0}(x-\frac{1}{2}a)$

$⟹x=\frac{a}{2}$

Now midpoint of $AS$ : $(\frac{0+a}{2},\frac{0+0}{2})$

$(\frac{a}{2},\frac{0}{2})$

If the equation of chords bisects the $AS$, it must pass through the midpoint of $AS$, which we can see clearly does so from equation $\left(3\right)$, hence proved.