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SIVARAMAKRISHNAN Vaithianathan Grade: Upto college level
        Find the locus of the point of intersection of tangents to the ellipse x^2/a^2 +y^2/b^2 = 1 at the points the sum of whose ordinates is constant(let it be c)
7 years ago

Answers : (1)

vikas askiitian expert
510 Points

eq of ellipse = x2/a2 + y2/b2 = 1              ............1

eq of tangent of this ellipse is y = mx +(a2m2+b2)1/2   .............2            (m is slope of tangent)

now we have to find point of intersection of this tangent & ellipse ,for that put eq 2 in 1


x2/a2 + [mx+(a2m2+b2)1/2]2/b2 = 1

x2(1/a2+m2/b2) + 2mx(a2m2+b2)1/2/b2 + (a2m2+b2)/b2  - 1 = 0

now it is given that the sum of ordinate is constant c , that means sum of roots of above equation is equal to c ...

x1+x2 = C                          (sum of roots = -b/a)

-[2m(a2m2+b2)1/2 /b2 ] / [1/a2+m2/b2] = C

squaring both sides & after simplifing we get

 m2 = C2b2/a2(4a2-c2)           ...................3

now ,  tangent will pass through point of intersection of tangents so let point of intersection is (A,B) then

this point will satisfy the eq of tangent so

 locus is  B = mA + (a2m2+b2)1/2                          (from eq 2)

  now substitute value of m in above eq & replace (A,B) by (X,Y) this will give the required locus.....

approve my ans if u like it

7 years ago
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