Question

The condition that the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, whose middle point is $(x_1,y_1)$ subtends a right angle at the centre of the ellipse, is:

• A. $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}+\frac{1}{b^2}){(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})}^2$
• B. $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}-\frac{1}{b^2}){(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})}^2$
• C. $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}+\frac{1}{b^2}){(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2})}^2$
• D. $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}-\frac{1}{b^2}){(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2})}^2$

Answer 1

The correct option is A $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}+\frac{1}{b^2}){(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})}^2$

Equation of chord is $T=S_1$
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}+\frac{{y_1}^2}{b^2}$
homogenising equation of ellipse w.r.t equation of chord
$\frac{x^2}{a^2}+\frac{y^2}{b^2}={\left(\frac{\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}}{\frac{x_1^2}{a^2}+\frac{{y_1}^2}{b^2}}\right)}^2$
Since, it represents a perpendicular pair of straight lines
then coefficient of $x^2=-$ coefficient of $y^2$
Therefore, $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}=(\frac{1}{a^2}+\frac{1}{b^2}){(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})}^2$