Question

Show that the tangents to the parabola $y^2=4ax$ at the ends of its latus rectum meet at its directrix.

Answer 1

take the parabola as $y^2=4ax$

now coordinates of its end of latus rectum are $(a,2a);(a,-2a)$.

Now tangent at any point $(x‘,y‘)$ on parabola is $yy‘=2a(x+x‘)$.

Tangent at point $(a,2a)$ is $2ay=2ax+2a^2$ and at $(a,-2a)$ is $-2ay=2ax+2a^2$.

By solving this linear equation in two unknown, we get point of intersection of tangent $y=0$ and $x=-a$;

so the point of intersection is $(-a,0)$ and it clearly lie on directrix.