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