Question

In the ellipse $\frac{x^2}{36}+\frac{y^2}{9}=1$, find the equation to the chord which passes through the point $(2,1)$ and is bisected at that point.

Answer 1

$\frac{x^2}{36}+\frac{y^2}{9}=1$

Mid point of chord is $(2,1)$

Equation of chord when mid point is given is $T=S^{\prime }$

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

$\frac{2x}{36}+\frac{y}{9}=\frac{4}{36}+\frac{1}{9}\frac{2x+4y}{36}=\frac{4+4}{36}2x+4y=8x+2y=4$

is the required equation of chord.