Question

find the locus of the mid-points of chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ that are parallel to the line $y=2x+c$

Answer 1

As we know that line parallel to line $y=2x+c$ is given by, $y=2x+k$ where k is a constant

Putting y = 2x+k in ellipse equation we get,

$\frac{x^2}{a^2}+\frac{(2x+k{)}^2}{b^2}=1$

$⟹x^2(b^2+4a^2)+4a^{2}kx+(k^2{a}^2-a^2{b}^2)=0$

from here we get, $x_1+x_2=-\frac{4a^{2}k}{b^2+4a^2}$

Mid point x-coordinate of chord is given by -$\frac{2a^{2}k}{b^2+4a^2}$

Similarily by substituting $x=\frac{y-k}{2}$

we get ,

$y^2(b^2+4a^2)-2b^{2}ky+(b^2{k}^2-4a^2{b}^2)=0$

Hence $\frac{y_1+y_2}{2}=\frac{b^{2}k}{b^2+4a^2}$

Hence, the locus of the midpoint of the chord is given by $y+2x=k$