InterviewSolution
Saved Bookmarks
| 1. |
`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`A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(12)` |
|
Answer» Correct Answer - B 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)` |
|