​​A chord PQ of the hyperbola xy = c² is tangent to the hyperbola x²/a² - y²/b² = -1. Find the locus of the middle point of PQ.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{xy}={\mathrm{c}}^2 \mathrm{is} \mathrm{a} \mathrm{transformed} \mathrm{form} \mathrm{of} {\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{where} \mathrm{c}=\frac{\mathrm{a}}{\sqrt{2}} \\ \Rightarrow \mathrm{a}=\mathrm{c}\sqrt{2} \\ \Rightarrow {\mathrm{a}}^2=2{\mathrm{c}}^2 \\ \mathrm{Hence} \mathrm{xy}={\mathrm{c}}^2 \mathrm{is} \mathrm{written} \mathrm{as} \\ {\mathrm{x}}^2-{\mathrm{y}}^2=2{\mathrm{c}}^2 \\ \mathrm{Let} \mathrm{mid} \mathrm{point} \mathrm{be} \left(\mathrm{h},\mathrm{k}\right), \mathrm{hence} \mathrm{equation} \mathrm{of} \mathrm{chord} \mathrm{is} \\ \mathrm{T}={\mathrm{S}}_1 \\ \mathrm{xh}-\mathrm{yk}={\mathrm{h}}^2-{\mathrm{k}}^2 \\ \Rightarrow \mathrm{yk}=\mathrm{xh}+\left({\mathrm{k}}^2-{\mathrm{h}}^2\right) \\ \Rightarrow \mathrm{y}=\frac{\mathrm{xh}}{\mathrm{k}}+\frac{\left({\mathrm{k}}^2-{\mathrm{h}}^2\right)}{\mathrm{k}} \\ \mathrm{Compare} \mathrm{with} \mathrm{y}=\mathrm{mx}+\mathrm{c} \\ \mathrm{m}=\frac{\mathrm{h}}{\mathrm{k}} \mathrm{and} \mathrm{c}=\frac{{\mathrm{k}}^2-{\mathrm{h}}^2}{\mathrm{k}} \\ \mathrm{Now} \mathrm{y}=\mathrm{mx} \mathrm{is} \mathrm{tangent} \mathrm{to} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1 \mathrm{is} \\ {\mathrm{c}}^2={\mathrm{b}}^2-{\mathrm{a}}^2{\mathrm{m}}^2 \\ \mathrm{Hence} \mathrm{locus} \mathrm{is} \\ {\left(\frac{{\mathrm{k}}^2-{\mathrm{h}}^2}{\mathrm{k}}\right)}^2={\mathrm{b}}^2-{\mathrm{a}}^2 \frac{{\mathrm{h}}^2}{{\mathrm{k}}^2} \\ \Rightarrow {\left({\mathrm{k}}^2-{\mathrm{h}}^2\right)}^2={\mathrm{k}}^2{\mathrm{b}}^2-{\mathrm{a}}^2{\mathrm{h}}^2 \\ \mathrm{h}\to \mathrm{x} \mathrm{and} \mathrm{k}\to \mathrm{y} \\ \mathbf{\Rightarrow }{\left({\mathbf{y}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\right)}^{\mathbf{2}}\mathbf{=}{\mathbf{y}}^{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}} \end{gathered}$$

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