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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

P=(acosθ,bsinθ) Q=(-asinθ,bcosθ) O=(0,0) L^2 =(x_2 −x_1 )^2 +(y_2 −y_1 )^2 (PO)^2 =a^2 cos^2 θ+b^2 sin^2 θ (OQ)^2 =a^2 sin^2 θ+b^2 cos^2 θ (PQ)^2 +(OQ)^2 =a^2 +b^2 b)A=(1/2)(x_1 y_2 −x_2 y_1 ) A=(1/2)(abcos^2 θ+absin^2 θ) A=(1/2)ab c) PQ(x,y)=(((x_1 +x_2 )/2),((y_1 +y_2 )/2)) ((2x)/a)=cosθ−sin θ ....(i) ((2y)/b)=sin θ+cos θ ....(ii) ((4x^2 )/a^2 )+((4y^2 )/b^2 )=2 ((2x^2 )/a^2 )+((2y^2 )/b^2 )=1

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

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