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The-normal-at-the-point-P-4cos-3sin-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-e




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.
ThenormalatthepointP(4cosθ,3sinθ)ontheellipsex216+y29=1meetsthexaxisandyaxisatAandBrespectivelyshowthatlocusofthemidpointofABisanellipsewiththesameeccentricityasgivenellipse.
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
eqnnormal(y3sinθ)=(dxdy)(4cosθ,3sinθ)(x4cosθ)x216+y29=12x×dxdy16+2y9=0x×dxdy8=2y9dxdy=16y9x=16×3sinθ9×4cosθ=43tanθ(y3sinθ)=4tanθ3(x4cosθ)putx=0y3sinθ=4sinθ3cosθ×4cosθy=3sinθ16sinθ3=7sinθ3B(0,7sinθ3)puty=03sinθ=4sinθ3cosθ(x4cosθ)x=9sinθcosθ4sinθ+4cosθx=9cosθ+16cosθ4=7cosθ4A(7cosθ4,0)MidpointofAB=(7cosθ8,7sinθ6)locusα=7cosθ8β=7sinθ6(8α7)2+(6β7)2=1locusx2(78)2+y2(76)2=1eccenrixity=(76)2(78)276=67×7×643682×62e2=67×7×18×6×27=74x216+y29=1e1=1694=74soe1=e2=74proved

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