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

The equation of the chord having $\left(x_1,y_1\right)$ as its middle point is

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\Rightarrow \frac{x{x}_1/a^2+y{y}_1/{b^2}_{ }}{x_1^2/a^2+y_1^2/b^2}=1$

The equation of the pair of lines joining the centre and the extremities of the chord is

$\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}={\left[\frac{x{x}_1/a^2+y{y}_1/b^2}{x_1^2/a^2+y_1^2/b^2}\right]}^2\\ \therefore {\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)}^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)={\left(\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}\right)}^2\end{array}$

Since these lines are perpendicular$\Rightarrow$ $\text{ coef. }{x}^2+\mathrm{coef}{y}^2=0$

$\Rightarrow \frac{1}{a^2}{\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b_2}\right)}^2-\frac{x_1^2}{a^4}+\frac{1}{b^2}{\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)}^2-\frac{y_1^2}{b^4}=0\Rightarrow \frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}={\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)}^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)$

$\therefore$The required condition is $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}={\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)}^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)$.