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Question Number 62753 by peter frank last updated on 24/Jun/19

The normal at the point P(4cos θ,3sin θ) on the ellipse (x^2 /(16)) +(y^2 /9)=1 meets the x−axis and y−axis at A and B respectively show that locus of the mid−point of AB is an ellipse with the same eccentricity as given ellipse.

$${The}\:{normal}\:{at}\:{the}\:{point} \\ $$$${P}\left(\mathrm{4cos}\:\theta,\mathrm{3sin}\:\theta\right)\:{on}\:{the} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{\mathrm{16}}\:+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:{meets} \\ $$$${the}\:{x}−{axis}\:{and}\:{y}−{axis} \\ $$$${at}\:{A}\:{and}\:{B}\:{respectively} \\ $$$${show}\:{that}\:{locus}\:{of}\:{the} \\ $$$${mid}−{point}\:{of}\:{AB}\:{is}\:{an} \\ $$$${ellipse}\:{with}\:{the}\:{same} \\ $$$${eccentricity}\:{as}\:{given} \\ $$$${ellipse}. \\ $$

Answered by Hope last updated on 25/Jun/19

eqn normal (y−3sinθ)=(((−dx)/dy))_((4cosθ,3sinθ)) (x−4cosθ) (x^2 /(16))+(y^2 /9)=1 ((2x×(dx/dy))/(16))+((2y)/9)=0 ((x×(dx/dy))/8)=((−2y)/9)→(dx/dy)=((−16y)/(9x))=((−16×3sinθ)/(9×4cosθ))=((−4)/3)tanθ (y−3sinθ)=((4tanθ)/3)(x−4cosθ) put x=0 y−3sinθ=((4sinθ)/(3cosθ))×−4cosθ y=3sinθ−((16sinθ)/3)=((−7sinθ)/3) B(0,((−7sinθ)/3)) put y=0 −3sinθ=((4sinθ)/(3cosθ))(x−4cosθ) x=((−9sinθcosθ)/(4sinθ))+4cosθ x=((−9cosθ+16cosθ)/4)=((7cosθ)/4)A(((7cosθ)/4),0) Mid point of AB=(((7cosθ)/8),((−7sinθ)/6)) locus α=((7cosθ)/8) β=((−7sinθ)/6) (((8α)/7))^2 +(((6β)/(−7)))^2 =1 locus (x^2 /(((7/8))^2 ))+(y^2 /(((7/6))^2 ))=1 eccenrixity=((√(((7/6))^2 −((7/8))^2 ))/(7/6))=(6/7)×7×(√((64−36)/(8^2 ×6^2 ))) e_2 =(6/7)×7×(1/(8×6))×2(√7) =((√7)/4) (x^2 /(16))+(y^2 /9)=1 e_1 =((√(16−9))/4)=((√7)/4) so e_1 =e_2 =((√7)/4)proved

$${eqn}\:{normal}\:\left({y}−\mathrm{3}{sin}\theta\right)=\left(\frac{−{dx}}{{dy}}\right)_{\left(\mathrm{4}{cos}\theta,\mathrm{3}{sin}\theta\right)} \:\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}{x}×\frac{{dx}}{{dy}}}{\mathrm{16}}+\frac{\mathrm{2}{y}}{\mathrm{9}}=\mathrm{0} \\ $$$$\frac{{x}×\frac{{dx}}{{dy}}}{\mathrm{8}}=\frac{−\mathrm{2}{y}}{\mathrm{9}}\rightarrow\frac{{dx}}{{dy}}=\frac{−\mathrm{16}{y}}{\mathrm{9}{x}}=\frac{−\mathrm{16}×\mathrm{3}{sin}\theta}{\mathrm{9}×\mathrm{4}{cos}\theta}=\frac{−\mathrm{4}}{\mathrm{3}}{tan}\theta \\ $$$$\left({y}−\mathrm{3}{sin}\theta\right)=\frac{\mathrm{4}{tan}\theta}{\mathrm{3}}\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$${put}\:{x}=\mathrm{0} \\ $$$${y}−\mathrm{3}{sin}\theta=\frac{\mathrm{4}{sin}\theta}{\mathrm{3}{cos}\theta}×−\mathrm{4}{cos}\theta \\ $$$${y}=\mathrm{3}{sin}\theta−\frac{\mathrm{16}{sin}\theta}{\mathrm{3}}=\frac{−\mathrm{7}{sin}\theta}{\mathrm{3}} \\ $$$${B}\left(\mathrm{0},\frac{−\mathrm{7}{sin}\theta}{\mathrm{3}}\right) \\ $$$${put}\:{y}=\mathrm{0} \\ $$$$−\mathrm{3}{sin}\theta=\frac{\mathrm{4}{sin}\theta}{\mathrm{3}{cos}\theta}\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$${x}=\frac{−\mathrm{9}{sin}\theta{cos}\theta}{\mathrm{4}{sin}\theta}+\mathrm{4}{cos}\theta \\ $$$${x}=\frac{−\mathrm{9}{cos}\theta+\mathrm{16}{cos}\theta}{\mathrm{4}}=\frac{\mathrm{7}{cos}\theta}{\mathrm{4}}\boldsymbol{{A}}\left(\frac{\mathrm{7}{cos}\theta}{\mathrm{4}},\mathrm{0}\right) \\ $$$${Mid}\:{point}\:{of}\:{A}\boldsymbol{{B}}=\left(\frac{\mathrm{7}{cos}\theta}{\mathrm{8}},\frac{−\mathrm{7}{sin}\theta}{\mathrm{6}}\right) \\ $$$${locus}\:\alpha=\frac{\mathrm{7}{cos}\theta}{\mathrm{8}}\:\:\:\beta=\frac{−\mathrm{7}{sin}\theta}{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{8}\alpha}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\beta}{−\mathrm{7}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${locus}\:\:\frac{{x}^{\mathrm{2}} }{\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\left(\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${eccenrixity}=\frac{\sqrt{\left(\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }}{\frac{\mathrm{7}}{\mathrm{6}}}=\frac{\mathrm{6}}{\mathrm{7}}×\mathrm{7}×\sqrt{\frac{\mathrm{64}−\mathrm{36}}{\mathrm{8}^{\mathrm{2}} ×\mathrm{6}^{\mathrm{2}} }} \\ $$$${e}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{7}}×\mathrm{7}×\frac{\mathrm{1}}{\mathrm{8}×\mathrm{6}}×\mathrm{2}\sqrt{\mathrm{7}}\:=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:\:{e}_{\mathrm{1}} =\frac{\sqrt{\mathrm{16}−\mathrm{9}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${so}\:{e}_{\mathrm{1}} ={e}_{\mathrm{2}} =\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}{proved} \\ $$$$ \\ $$$$ \\ $$

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