Transform the equation x/a + y/b = 1 into the normal form when a>0, b>0. If the perpendicular distance of the straight line from the origin is P, deduce that 1/p² = 1/a² + 1/b²​.

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$$\begin{gathered} \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1....\left(\mathrm{i}\right), \mathrm{a}>0, \mathrm{b}>0 \\ \sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2}}=\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}} \\ \mathrm{Divide} \left(\mathrm{i}\right) \mathrm{through} \mathrm{out} \mathrm{by} \frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{a}.\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}}}+\frac{\mathrm{y}}{\mathrm{b}.\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}}}=\frac{1}{\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}}} \\ \Rightarrow \mathrm{x}\left(\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)+\mathrm{y}\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}....\left(\mathrm{i}\right) \\ \mathrm{Slope} \mathrm{of} \left(\mathrm{i}\right) \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{tan\alpha }=-\frac{\mathrm{Ceofficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}} \\ =-\frac{{\displaystyle \frac{1}{\mathrm{a}}}}{{\displaystyle \frac{1}{\mathrm{b}}}}=-\frac{\mathrm{b}}{\mathrm{a}} \mathrm{which} \mathrm{is} \mathrm{negative}, \mathrm{hence} \mathrm{\alpha } \mathrm{is} \mathrm{obtuse}. \\ \mathrm{tan\alpha }=-\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{Perendicular} }{\mathrm{Base}} \\ \mathrm{Hypotenuse}=\sqrt{{\mathrm{Base}}^2+{\mathrm{Perpendicular}}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \mathrm{For} \mathrm{obtuse} \mathrm{angles} \sin \mathrm{is} \mathrm{positive} \mathrm{and} \cos \mathrm{in} \mathrm{negative}. \\ \mathrm{sin\alpha }=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{cos\alpha }=-\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes} \\ \mathrm{x}\left(-\mathrm{cos\alpha }\right)+\mathrm{y}\left(\mathrm{sin\alpha }\right)=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \sin \left(\mathrm{A}\right)=\sin \left(\mathrm{\pi }-\mathrm{A}\right) \mathrm{and} \cos \left(\mathrm{A}\right)=-\cos \left(\mathrm{\pi }-\mathrm{A}\right) \\ \Rightarrow \mathrm{xcos}\left(\mathrm{\pi }-\mathrm{\alpha }\right)+\mathrm{ysin}\left(\mathrm{\pi }-\mathrm{\alpha }\right)=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{which} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{normal} \mathrm{form} \mathrm{where} \mathrm{tan\alpha } \mathrm{is} \mathrm{slope} \mathrm{of} \mathrm{given} \mathrm{line}. \\ \mathrm{Hence} \mathrm{right} \mathrm{hand} \mathrm{side} \mathrm{must} \mathrm{give} \mathrm{perpendicular} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{straight} \mathrm{line} \mathrm{from} \mathrm{the} \mathrm{origin}. \\ \Rightarrow \mathrm{p}=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \Rightarrow {\mathrm{p}}^2=\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2} \\ \Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2}+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2} \\ \mathbf{\Rightarrow }\frac{\mathbf{1}}{{\mathbf{p}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{b}}^{\mathbf{2}}} \end{gathered}$$

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