$\text{ The locus of poles of chords of the hyperbola }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\text{ which subtend a right angle at the centre is }$

$$\begin{gathered} \text{ Given hyperbola }\frac{x^2}{a^2}\text{-}\frac{y^2}{b^2}\text{=1} \\ \text{Equation of polar is }\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=1....\left(1\right) \\ \Rightarrow \text{ Homogenising hyperbola }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 by \mathrm{using} \left(1\right) \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1{(\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2})}^2 \\ then by hypothesis, \\ \mathrm{coef}\text{ of }{\mathrm{x}}^2+\mathrm{coef}\text{ of }{\mathrm{y}}^2=0 \\ \Rightarrow (\frac{1}{a^2}-\frac{{x^2}_1}{a^4})-(\frac{1}{b^2}-\frac{{y^2}_1}{b^4})=0 \end{gathered}$$

locus is $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^4}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^4}=\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}$