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AP EAMCET201725 Apr 2017Evening ShiftMathematicsEllipseActual

The equation of the locus of the foot of the perpendicular drawn from the centre of the ellipse x^2 a^2 + y^2 b^2 =1 to any tangent of the ellipse is

Options

  1. A(x^2+y^2 )^2=a^2 x^2+b^2 y^2
  2. B(x^2-y^2 )^2=a^2 x^2+b^2 y^2
  3. C(x^2+y^2 )^2=a^2 x^2-b^2 y^2
  4. D(x^2-y^2 )^2=a^2 x^2-b^2 y^2

Correct answer

A. (x^2+y^2 )^2=a^2 x^2+b^2 y^2

Step-by-step solution

No solution. Refer to answer key.

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