Prove that the product of perpendicular drawn from foci to any tangent of the ellipse  x²/a² +y²/b² is b²

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{Any} \mathrm{tangent} \mathrm{to} \mathrm{ellipse} \mathrm{is} \\ \mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2} \\ \Rightarrow \mathrm{mx}-\mathrm{y}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}=0 \\ \mathrm{If} {\mathrm{p}}_1 \mathrm{and} {\mathrm{p}}_2 \mathrm{be} \mathrm{lengths} \mathrm{of} \mathrm{perpendicular} \mathrm{from} \mathrm{foci} {\mathrm{S}}_1\left(\mathrm{ae},0\right) \mathrm{and} {\mathrm{S}}_2\left(-\mathrm{ae},0\right) \mathrm{respectively} \\ {\mathrm{p}}_1=\frac{\left|\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}+\mathrm{aem}\right|}{\sqrt{1+{\mathrm{m}}^2}} \\ {\mathrm{p}}_2=\frac{\left|\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}-\mathrm{aem}\right|}{\sqrt{1+{\mathrm{m}}^2}} \\ {\mathrm{p}}_1.{\mathrm{p}}_2=\frac{\left|\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}+\mathrm{aem}\right|}{\sqrt{1+{\mathrm{m}}^2}}.\frac{\left|\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}-\mathrm{aem}\right|}{\sqrt{1+{\mathrm{m}}^2}} \\ =\frac{\left|{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2{\mathrm{e}}^2{\mathrm{m}}^2\right|}{1+{\mathrm{m}}^2} \\ =\frac{\left|{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2{\mathrm{e}}^2{\mathrm{m}}^2\right|}{1+{\mathrm{m}}^2} \\ =\frac{\left|{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2-\left({\mathrm{a}}^2-{\mathrm{b}}^2\right){\mathrm{m}}^2\right|}{1+{\mathrm{m}}^2} \left[\mathrm{As} {\mathrm{a}}^2{\mathrm{e}}^2=\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)\right] \\ =\frac{\left|{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2{\mathrm{m}}^2\right|}{1+{\mathrm{m}}^2} \\ =\frac{\left|{\mathrm{b}}^2+{\mathrm{b}}^2{\mathrm{m}}^2\right|}{1+{\mathrm{m}}^2} \\ ={\mathrm{b}}^2\left(\frac{1+{\mathrm{m}}^2}{1+{\mathrm{m}}^2}\right) \\ ={\mathrm{b}}^2 \end{gathered}$$

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