Question

Find the equation of the circle, one of whose diameters is the latus rectum $y^2=4ax$. Show that this circle goes through the common point of the axis and the directrix of the parabola.

Answer 1

Latus rectum and points of $y^2=4ax$

$arc(a,2a)$ & $(a-2a)$

centre $=(a,0)$

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

Circle $(x-a{)}^2+y^2=4a^2$

It also passes through $(-a,0)$

$(-a-a{)}^2+0^2=4a^2$

$4a^2=4a^2$

$(-a,0)$ is intersection point of direction & axis of parabola