If circumcentre of an equilateral triangle inscribed in `x^(2)/a^(2) + y^(2)b^(2) = 1,` with vertices having eccentric angles `alpna, beta, gamma,` respectively is `(x_(1), y_(1))` then `sum cos alpha cos beta + sum sin alpha sin beta =`

Options

A. `(9x_(1)^(2))/(a^(2)) + (9y_(1)^(2))/b^(2) + 3/2`
B. `9x_(1)^(2) - 9 y_(1)^(2) + a^(2)b^(2)`
C. `(9x_(1)^(2))/(a) + (9y_(1)^(2))/(b) + 3`
D. `(9x_(1)^(2))/(2a^(2)) + (9y_(1)^(2))/(2b^(2)) - 3/2`

Answer

D

Explanation

Let `P (a cos alpha, b sin alpha), Q (a cos beta, b sin beta)` and `R (a cos gamma, b sin gamma)` be the vertices of equilateral triangle PQR.
Then,
`x_(1) = a/3(a cos alpha = cos beta + cos gamma), y_(1) = b/3(sin alpha + sin beta + sin gamma)`
`rArr ((3x_(1))/(a))^(2) + ((3y_(1))/(b))^(2) = (cos alpha + cos beta + cos gamma)^(2) + (sin alpha + sin beta = sin gamma)^(2)`
`rArr (9x_(1)^(2))/(a^(2)) + (9y_(1)^(2))/(b^(2) = 3 + 2 (sum cos alpha cos beta + sum sin alpha sin beta)`
`rArr sum cos alpha cos beta + sum sin alpha sin beta = (9x_(1)^(2))/(2a^(2)) + (9y_(1)^(2))/(2b^(2)) - 3/2`