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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» 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` |
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