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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,

$$y_1^2-4ax_1 = yy_1 -2a(x+x_1)$$

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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 $T=S_1$.

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  • $\begingroup$ In the second step, how did you get $ \frac{y_a + y_b}{2 }=2𝑎𝑘$? $\endgroup$
    – Ardent
    Nov 21 '19 at 21:18
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    $\begingroup$ @Kaushik - The sum of the two roots of the equation $y^2-4aky-4am=0$ is 4ak, which leads to $(y_a+y_b)/2 = 2ak$. $\endgroup$
    – Quanto
    Nov 21 '19 at 21:24
  • $\begingroup$ What about this - $x_1= \frac1{2a}(y_a^2+y_b^2)$? How did you get this one? $\endgroup$
    – Ardent
    Nov 22 '19 at 5:46
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    $\begingroup$ @Kaushik - Use $y^2=4ax$ to write $x_1= \frac 12 (x_a+x_b)= \frac1{8a}(y_a^2+y_b^2)$. There was a typo. Thanks $\endgroup$
    – Quanto
    Nov 22 '19 at 13:23
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    $\begingroup$ @Kaushik - It'd work, but it'd be a little awkward since the parabola is given as $y^2=4ax$. $\endgroup$
    – Quanto
    Nov 22 '19 at 19:42
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The slope of a chord of parabola $y^2=4ax$ is $2a/y_1$, where $y_1$ is the ordinate of its midpoint. This is a property you can easily check and it can also be stated as follows: the slope of the chord is the same as the lope of the tangent at the point on the parabola with ordinate $y_1$. The equation of the chord is then: $$ y-y_1={2a\over y_1}(x-x_1) $$ which is equivalent to the given equation.

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