`P and Q` are points on the ellipse `x^2/a^2+y^2/b^2 =1` whose center is `C`. The eccentric angles of P and Q differ by a right angle. If `/_PCQ` minimum, the eccentric angle of P can be (A) `pi/6` (B) `pi/4` (C) `pi/3` (D) `pi/12`

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(12)`

Answer

B

Explanation

Since the eccentric angles of P and Q differ by a right angle, we can take P as `(a cos theta, b sin theta)` and Q as `(-a sin theta, b cos theta)`
Slope of `CP = (b sin theta)/(a cos theta)`
Slope of `CQ =-(b cos theta)/(a sin theta)`
If Q is the angle between CP and CQ
`A = tan^(-1) |((bsin theta)/(a cos theta)+(b cos theta)/(a sin theta))/(1-(b^(2)sin theta cos theta)/(a^(2)cos theta sin theta))| = tan^(-1) |(2ab)/(a^(2)-b^(2))-(1)/(sin 2 theta)|`
Q is minimum if `sin 2 theta` is maximum.
i.e., if `2 theta = (pi)/(2)` or `theta = (pi)/(4)`