P and Q are corresponding points on an ellipse and the auxiliary circle respectively. The normal at P to the ellipse meets CQ in R .......?

where C is the center of the ellipse. Find CR ?

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  • 7 months ago
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    Let P be (a cosθ, b sinθ ) so that Q is ( a cosθ, a sinθ ) which lies on the auxiliary circle, C is (0,0)

    Equation of CQ is (y-0) = (a sin⁡θ)/(a cos⁡θ ) (x-0)

    Or y = tan θ ……………………… (1)

    Equation of normal at P is ax/cos⁡θ = by/sin⁡θ = a^2-b^2 …………. (2)

    Both (1) and (2) intersect at R. Solving (1) and (2) we get -

    ax/(cos⁡θ)= b/sinθ * x tan θ = a^2-b^2

    Hence x = ( a + b ) cos θ and y = x tan θ = (a+b) sin θ

    R is [ (a+b) cos θ ; (a+b) sin θ ] and C is (0,0)

    Therefore CR^2 = (a+b)^2 (cos^2 θ + sin^2 θ ) = (a+b)^2

    Therefore CR = ( a + b ) ……………………… Answer

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