# Find the locus of the midpoints of the chords of the ellipse that are parallel to y = 2x+c

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?

• 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. – conditionalMethod Nov 7 '19 at 19:11
• @conditionalMethod Strictly speaking, the locus is a line segment. – amd Nov 8 '19 at 6:51

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 $$x(b^2)+y(2a^2)=0.$$

• Oops I got there but did a silly miscalculation in the expansion. – Techie5879 Nov 7 '19 at 20:02

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 $$2a^2 y + b^2 x = 0$$

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 $$c$$