Question

The eccentric angles of the ends of latusrectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is:

• A. ${\tan }^{-1}(\pm \frac{b}{ae})$
• B. ${\sin }^{-1}(\pm \frac{b}{ae})$
• C. ${\cos }^{-1}(\pm \frac{b}{ae})$
• D. ${\sec }^{-1}(\pm \frac{b}{ae})$

Answer 1

Answer: (A)

${\tan }^{-1}(\pm \frac{b}{ae})$

Let , $\theta$ be the eccentric Angle of an end of a latus rectum. Then, the coordinates of the end of latus rectum is
$(acos\theta ,bsin\theta )$
As , we know that the coordinates of latus rectum is
$(ae,\pm \frac{b^2}{a})$
$acos\theta =ae$
$cos\theta =e$
$bsin\theta =\pm \frac{b^2}{a}$
$sin\theta =\pm \frac{b}{a}$
$\therefore tan\theta =\pm \frac{b}{ae}$
$\therefore \theta =ta{n}^{-1}(\pm \frac{b}{ae})$
Hence , option A