The locus of the foot of perpendicular drawn from the centre of the ellipse x 2 + 3 y 2 = 6 on any tangent to it is

Step-by-step solution

We have, x 2 + 3 y 2 = 6 ⇒ x 2 6 + y 2 2 = 1 Centre of an ellipse is 0 , 0 . Let the foot of perpendicular be ( h , k ) . Equation of tangent in slope form to the standard ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 . Hence, equation of tangent with slope m passing ( h , k ) is y = m x ± 6 m 2 + 2 ⇒ k = m h ± 6 m 2 + 2 , where m = - h k . ⇒ k = h - h k ± 6 m 2 + 2 ⇒ 6 h 2 k 2 + 2 = h 2 + k 2 k ⇒ 6 h 2 + 2 k 2 = h 2 + k 2 2 So, required locus is 6 x 2 +