The circumcenter of the equilateral triangle having the three points $\theta_1, \theta_2, \theta_3$ lying on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its vertices is $(r, s)$. Then the average of $\cos \left(\theta_1-\theta_2\right)$, $\cos \left(\theta_2-\theta_3\right)$ and $\cos \left(\theta_3-\theta_1\right)$ is

• Correct answer
  A

  $\frac{1}{2}\left[\frac{3 r^2}{a^2}+\frac{3 s^2}{b^2}-1\right]$
• B

  $\frac{3}{2}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}\right]$
• C

  $\frac{1}{3}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}\right]$
• D

  $\frac{1}{3}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}+\frac{r s}{a b}\right]$

Hence, the correct answer is option: 'A'.