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I am asked to find the locus of the midpoints of the chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ that are parallel to the line $y = 2x+c$.

So I clearly get that the slope of the chord must be equal to $2$

Now, i know that for an ellipse, the equation of the chord that is bisected at $(x_1,y_1)$ is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}$$

But here i get two extra variables $x,y$ other than $x_1,y_1$ (of which I need to find the locus).

So how do I go about doing this?

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  • $\begingroup$ The chords have a limit position when the line becomes tangent instead of intersecting the ellipse at two points or are none. The corresponding mid-points will coincide with those tangency points. Two points determine a line. Prove that the locus must be a line, compute the two tangency points and then the line passing through them. $\endgroup$ – conditionalMethod Nov 7 '19 at 19:11
  • $\begingroup$ @conditionalMethod Strictly speaking, the locus is a line segment. $\endgroup$ – amd Nov 8 '19 at 6:51
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From the equation of the chords bisected at point $(h,k)$ (instead of $x_1, y_1$ I am using $h,k$ to avoid typing subscripts) we can get the slope of the chord as $-\frac{hb^2}{ka^2}$. But this should be equal to the slope of the given line. So $$-\frac{hb^2}{ka^2}=2 \implies h(b^2)+k(2a^2)=0.$$ So the locus is $$x(b^2)+y(2a^2)=0.$$

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  • $\begingroup$ Oops I got there but did a silly miscalculation in the expansion. $\endgroup$ – Techie5879 Nov 7 '19 at 20:02
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If we say $u = \frac {x}{a}, v = \frac {y}{b}$

Then we have the question what is the locus of points such that

$bv = 2au + c$

Intersects

$u^2 + v^2 = 1$

Circles are just easier to work with.

We know that the midpoints will be on the line $2av + bu = 0$ i.e. perpendicular to the chord line and through the origin.

And now we transform back to our original system.

$\frac {2a}{b} y + \frac {b}{a}x = 0$ or $2a^2 y + b^2 x = 0$

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Let us find the abcissa of the two possible values of intersection

$$0=b^2x^2+a^2(2x+c)^2-a^2b^2=x^2(4a^2+b^2)-4a^2cx+()$$

So, if $M(h,k)$ is the midpoint, $$h=\dfrac{4ca^2}{2(4a^2+b^2)}$$

Similarly, $$k=c+2\cdot\dfrac{4ca^2}{2(4a^2+b^2)}$$

Replace the value of $c$ from the first equation in to the second one to eliminate $c$

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