Question

If in the ellipse $\frac{x^2}{36}+\frac{y^2}{9}=1$, a chord through the points $(4,0)$ has the points $(2,\alpha )$ as its mid point, find $\alpha$.

Answer 1

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

Equation of the chord with given mid point:

$T=S1$

Mid point $(2,\alpha )$

$⟹\frac{2x}{36}+\frac{\alpha y}{9}-1=\frac{4}{36}+\frac{{\alpha }^2}{9}-1$

$⟹\frac{x}{18}+\frac{\alpha y}{9}=\frac{1+{\alpha }^2}{9}$ (pass from $(4,0)$)

$⟹\frac{4}{18}+\frac{\alpha \left(0\right)}{9}=\frac{1+{\alpha }^2}{9}$

$⟹\frac{2}{9}=\frac{(1+{\alpha }^2)}{9}$

$⟹{\alpha }^2=1$

$⟹\alpha =\pm 1$