Question

The locus of the point of intersection of tangents of `x^2/a^2 + y^2/b^2 =1` at two points whose eccentric angles differ by `pi/2` is an ellipse whose semi-axes are: (A) `sqrt2 a,sqrt 2 b` (B) `a/sqrt2,b/sqrt2` (C) `a/2,b/2` (D) `2a,2b`

Answer 1

Let the points on ellipse be `p(acos(theta),bsin(theta))` and
`Q(acos(pi/2+theta),bsin(pi/2+theta))=(-asintheta,bcostheta)`
Equation of tangent at point p:
`x/acostheta+y/bsintheta=1`
Equation of tangent at point Q:
`-x/asintheta+y/bcostheta=1`
On solving these equations we get, `x/a=(costheta-sintheta)`
`y/b=(costheta+sintheta)`
On squaring and adding these equations,
`2=(x/a)^2+ (y/b)^2` Therefore,
`(x/(asqrt2))^2+(y/(bsqrt2))^2=1`