Finding the equation of chord of contact to a parabola when midpoint of the chord is given.
Let $y^2=4ax$ be the parabola, $P(x_1,y_1)$ be the midpoint, and
$$S_1 = y_1^2 - 4ax_1,\>\>\>\>\>\>\>\>T = yy_1 + 2a(x+x_1)$$
In my text book it is given that equating $T=S_1$ gives the equation of the chord of contact, but I don't get how this thing works.
An explanation on this would help me a lot.
My attempt:
Let the equation of chord of contact be $\frac{y - c}{m} = x$.
Then, $$ y^2 - \frac{4ay}{m} + \frac{4ac}{m} = 0$$
$$2y_1 = \frac{4a}{m} \implies m = \frac{2a}{y_1}$$ Now, $$ x_1 = \frac{x_a + x_b}{2}$$ $$x_1 = \frac{(y_a + y_b)^2 - 2y_ay_b}{8a}$$ By Substituting we get, $$c = y_1 - \frac{2ax_1}{y_1}$$
So, the equation is
$$y-y_1+\frac{2ax_1}{y_1} = \frac{2a}{y_1}x\implies yy_1-y_1^2+{2ax_1} = {2a}x $$
Rearrange to get,
$$y_1^2-4ax_1 = yy_1 -2a(x+x_1)$$