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,
$$y_1^2-4ax_1 = yy_1 -2a(x+x_1)$$
Best Answer
Let $x=ky+m$ be the equation of the chord. The $y$-coordinates of the two intersections $y_a$ and $y_b$ are given by
$$ y^2-4aky-4am=0$$
Then, relate to the midpoint of the chord to get,
$$\frac{y_a+y_b}2= 2ak=y_1,\>\>\>\>\>\frac{x_a+x_b}2 =x_1$$
where the first equation yields $k=\frac{y_1}{2a}$. Reexpress the second equation,
$$x_1= \frac1{8a}(y_a^2+y_b^2)=\frac1{8a}[(y_a+y_b)^2-2y_ay_b]=2ak^2+m$$
which yields $m=x_1-\frac{y_1^2}{2a}$. Thus, the equation of the chord is $x= \frac{y_1}{2a}y+x_1-\frac{y_1^2}{2a}$, or
$$y_1^2-4ax_1=yy_1-2a(x+x_1)$$
which is $T=S_1$.