Circle drawn with chord of parabola as diameter touches the parabola

Question

I'm trying to prove that if the difference of ordinates of endpoints of the chord of a parabola is twice the length of its latus rectum, then the circle drawn with this chord as the diameter touches the parabola.
Not being too rigorous, I considered the parabola $y^2=4ax$ and the circle $x^2+y^2+2gx+2fy+c=0$
Any general point on $y^2=4ax$ can be given as $(at^2,2at)$, so I plugged this point into my circle's equation. I got
$a^2t^4+(4a^2+2ag)t^2+4aft+c=0$
If I consider $t_1, t_2, t_3, t_4$ to be the roots of the above equation, I get
$t_1+t_2+t_3+t_4=0$ - (i)
$t_1t_2t_3+t_2t_3t_4+t_1t_3t_4+t_1t_2t_4=4+\frac{2g}{a}$ - (ii)
$t_1t_2+t_2t_3+t_3t_4+t_1t_4+t_1t_3+t_2t_4=-\frac{4f}{a}$ - (iii)
$t_1t_2t_3t_4=\frac{c}{a^2}$ - (iv)
For any two points $(at_1^2,2at_1)$ and $(at_2^2,2at_2)$, the difference of ordinates is $2a(t_1-t_2)$
Also the length of latus rectum for out parabola is $4a$
Since it states that the difference of the ordinates is equal to twice the length of the latus rectum,
$2a(t_1-t_2)=8a$ $\implies$ $t_1-t_2=4$ - (v)
But I'm unable to figure out how to prove that the circle will touch the parabola using the relations (i),(ii),(iii),(iv) and (v) above. Please show me a way ahead or a better way to prove it.

Answer 1

First of all we can simplify things a bit by setting, without loss of generality, $a=1/4$, so that the equation of the parabola is $x=y^2$ and latus rectum is $1$. But of course you are free to repeat the reasoning below with a more general equation.

Take then two points $A$ and $B$ on your parabola whose ordinates differ by $2$ and compute their midpoint $C$ (the centre of the circle of diameter $AB$): $$ A=(y^2,y),\quad B=((y+2)^2,y+2),\quad C=(y^2+2y+2,y+1). $$ Note that point $S$ on the parabola with the same ordinate as $C$ is $$ S=((y+1)^2,y+1), $$ while its symmetric $T$ about $x$-axis (which is also on the parabola) is $$ T=((y+1)^2,-y-1). $$ The tangent to the parabola at $T$ has a slope which is the opposite of the slope of the tangent at $S$, which in turn has the same slope as $AB$ (by a well-known property of parabolas). Hence you can easily check that $CT$ is perpendicular to the tangent at $T$.

Finally you can check that $CT=AC$, so that $T$ is a point on the circle. And the circle is thus tangent to the parabola at $T$.