Question

The eccentric angles of the extremities of latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are given by:

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

Answer 1

The correct option is C ${\tan }^{-1}(\pm \frac{b}{ae})$

Latus rectum is perpendicular to the minor axis and passes through the focus.

X- co-ordinate of latus rectum $=ae$

Equation of ellipse:$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Putting the value $x=ae$ we get:

$\frac{{\left(ae\right)}^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow e^2+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{y^2}{b^2}=1-e^2$

$\Rightarrow \frac{y^2}{b^2}=\frac{b^2}{a^2}$ $\Rightarrow y=\frac{b^2}{a}$

Extrimities are $(ae,\frac{b^2}{a})$ and $(ae,-\frac{b^2}{a})$

$\therefore \tan \theta =\pm \frac{b}{ae}$

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