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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)`
6 years ago

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
```
6 years ago
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