If S₁N₁ and S₂N₂ b e the perpendiculars from the foci S₁ and S₂ of an ellipse upon any tangent to the ellipse then prove that N₁ and N₂ lie on the auxillary circle  and S₁N₁ :S₂N₂ = b².
 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{to} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \mathrm{is} \\ \frac{\mathrm{x}}{\mathrm{a}}\mathrm{cos\theta }+\frac{\mathrm{y}}{\mathrm{b}}\mathrm{sin\theta }=1....\left(\mathrm{i}\right) \\ {\mathrm{S}}_1\left(\mathrm{ae},0\right) \mathrm{and} {\mathrm{S}}_2\left(\mathrm{ae}\text{'},0\right) \mathrm{are} \mathrm{foci}. \\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{perpendicular} \mathrm{to} \left(\mathrm{i}\right) \mathrm{and} \mathrm{passing} \mathrm{through} \left(\mathrm{ae},0\right) \mathrm{is} \\ \frac{\mathrm{x}}{\mathrm{b}}\mathrm{sin\theta }-\frac{\mathrm{y}}{\mathrm{a}}\mathrm{cos\theta }=\mathrm{k} \\ \mathrm{Put} \left(\mathrm{ae},0\right) \\ \frac{\mathrm{ae}}{\mathrm{b}}\mathrm{sin\theta }=\mathrm{k} \\ \mathrm{Hence} \mathrm{equation} \mathrm{becomes} \\ \frac{\mathrm{x}}{\mathrm{b}}\mathrm{sin\theta }-\frac{\mathrm{y}}{\mathrm{a}}\mathrm{cos\theta }=\frac{\mathrm{ae}}{\mathrm{b}}\mathrm{sin\theta }.....\left(\mathrm{ii}\right) \\ {\left(\mathrm{i}\right)}^2+{\left(\mathrm{ii}\right)}^2 \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}{\cos }^2\mathrm{\theta }+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}{\sin }^2\mathrm{\theta }+2\frac{\mathrm{xy}}{\mathrm{ab}}\mathrm{sin\theta }.\mathrm{cos\theta }+\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}{\sin }^2\mathrm{\theta }+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}{\cos }^2\mathrm{\theta }-2\frac{\mathrm{xy}}{\mathrm{ab}}\mathrm{sin\theta }.\mathrm{cos\theta }=1+\frac{{\mathrm{a}}^2{\mathrm{e}}^2}{{\mathrm{b}}^2}{\sin }^2\mathrm{\theta } \\ \Rightarrow {\mathrm{x}}^2\left(\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}\right)+{\mathrm{y}}^2\left(\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}\right)=\frac{{\mathrm{b}}^2+\left({\mathrm{a}}^2-{\mathrm{b}}^2\right){\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2} \\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\left(\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}\right)=\frac{{\mathrm{b}}^2\left(1-{\sin }^2\mathrm{\theta }\right)+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2} \\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\left(\frac{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2{\mathrm{b}}^2}\right)=\frac{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{Hence} \mathrm{N} \mathrm{and} {\mathrm{N}}_1 \mathrm{lie} \mathrm{on} \mathrm{auxillary} \mathrm{circle}. \\ \mathrm{Further} \\ {\mathrm{S}}_1{\mathrm{N}}_1.{\mathrm{S}}_2{\mathrm{N}}_2=\frac{\left(1-\mathrm{ecos\theta }\right)}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\times \frac{\left(1+\mathrm{ecos\theta }\right)}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}}} \\ =\frac{1-{\mathrm{e}}^2{\cos }^2\mathrm{\theta }}{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}} \\ =\frac{1-\left({\displaystyle \frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}}\right){\cos }^2\mathrm{\theta }}{\frac{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2{\mathrm{b}}^2}} \\ ={\mathrm{b}}^2\left(\frac{{\mathrm{a}}^2-{\mathrm{a}}^2{\cos }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}\right) \\ ={\mathrm{b}}^2\left(\frac{{\mathrm{a}}^2{\sin }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}\right) \\ {\mathbf{S}}_{\mathbf{1}}{\mathbf{N}}_{\mathbf{1}}\mathbf{.}{\mathbf{S}}_{\mathbf{2}}{\mathbf{N}}_{\mathbf{2}}\mathbf{=}{\mathbf{b}}^{\mathbf{2}}\mathbf{ } \end{gathered}$$


Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards