For t ∈ 0 , 2 π , if A B C is an equilateral triangle with vertices A sin t , - cos t , B cos t , sin t and C a , b such that its orthocentre lies on a circle with centre 1 , 1 3 , then a 2 - b 2 is equal to

Step-by-step solution

We know that for an equilateral triangle the orthocentre and centroid coincide. Here, centroid h , k ≡ cos t + sin t + a 3 , sin t - cos t + b 3 ⇒ 3 h - a = cos t + sin t                 . . . i ⇒ 3 k - b = sin t - cos t           . . . ii Eliminating t from above two equation i   &   ii , we get h - a 3 2 + k - b 3 2 = 2 9 So, h , k lies on the circle whose centre is a 3 , b 3 , now comparing with 1 , 1 3 we get, a = 3