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Question Number 50047 by peter frank last updated on 13/Dec/18

Answered by peter frank last updated on 13/Dec/18

$$\mathrm{P}=(\mathrm{acos}\theta ,\mathrm{bsin}\theta )$$$$\mathrm{Q}=(-\mathrm{asin}\theta ,\mathrm{bcos}\theta )$$$$\mathrm{O}=(0,0)$$$${\mathrm{L}}^2={({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2$$$${\left(\mathrm{PO}\right)}^2={\mathrm{a}}^2\cos {}^2\theta +{\mathrm{b}}^2\sin {}^2\theta$$$${\left(\mathrm{OQ}\right)}^2={\mathrm{a}}^2{\sin }^2\theta +{\mathrm{b}}^2{\cos }^2\theta$$$${\left(\mathrm{PQ}\right)}^2+{\left(\mathrm{OQ}\right)}^2=a^2+b^2$$$$$$$$b)\mathrm{A}=\frac{1}{2}({\mathrm{x}}_1{\mathrm{y}}_2-{\mathrm{x}}_2{\mathrm{y}}_1)$$$$\mathrm{A}=\frac{1}{2}(\mathrm{abcos}{}^2\theta +\mathrm{absin}{}^2\theta )$$$$\mathrm{A}=\frac{1}{2}\mathrm{ab}$$$$\mathrm{c})\mathrm{PQ}(\mathrm{x},\mathrm{y})=(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2})$$$$\frac{2\mathrm{x}}{\mathrm{a}}=\cos \theta -\sin \theta ....\left(\mathrm{i}\right)\frac{2\mathrm{y}}{\mathrm{b}}=\sin \theta +\cos \theta ....\left(\mathrm{ii}\right)$$$$\frac{{4\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{4\mathrm{y}}^2}{{\mathrm{b}}^2}=2$$$$\frac{{2\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{2\mathrm{y}}^2}{{\mathrm{b}}^2}=1$$$$$$

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