Question

Find the condition on a and b for which two distinct chords of the ellipse $\frac{x^2}{2a^2}+\frac{y^2}{2b^2}=1$ passing
through $(a,-b)$ are bisected by the line $x+y=b.$

Answer 1

Let the line $x+y=b$ bisect the chord at P$(a,b-a)$

$\therefore$ equation of chord whose mid-point is P $(a,b-a)\frac{xa}{2a^2}+\frac{y(b-a)}{2b^2}=\frac{a^2}{2a^2}+\frac{{(b-a)}^2}{2b^2}$

Since it passes through (a,-b) $\therefore \frac{a}{2a}-\frac{(b-a)}{2b}=\frac{a^2}{2b}+\frac{{(b-a)}^2}{2b^2}\Rightarrow (\frac{1}{a}+\frac{1}{b})a-1=a^2(\frac{1}{a^2}+\frac{1}{b^2})-\frac{2}{b}a+1\Rightarrow a^2(\frac{1}{a^2}+\frac{1}{b_2})-(\frac{3}{b}+\frac{1}{a})a+2=0$

Since line bisect two chord $\therefore$ above quadratic equation in a must have two distinct real roots

$\therefore {(\frac{3}{b}+\frac{1}{a})}^2-4(\frac{1}{a^2}+\frac{1}{b^2}).2>0\Rightarrow \frac{9}{b^2}+\frac{1}{a^2}+\frac{6}{ab}-\frac{8}{a^2}-\frac{8}{b^2}>0\Rightarrow \frac{1}{b^2}-\frac{7}{b^2}+\frac{6}{ab}>0\Rightarrow a^2-7b^2+6ab>0\Rightarrow a^2>7b^2-6ab$.