Question

The eccentric angle of a point on the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at a distance of $\frac{5}{4}$ units from the focus on the positive x-axis, is

• A. ${\cos }^{-1}\left(\frac{3}{4}\right)$
• B. $\pi -{\cos }^{-1}\left(\frac{3}{4}\right)$
• C. $\pi +{\cos }^{-1}\left(\frac{3}{4}\right)$
• D. None of these

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{3}{4}\right)$

Any point on the ellipse is $(a\cos \theta ,b\sin \theta )$.

i.e. $(2\cos \theta ,\sqrt{3}\sin \theta )$.
The focus on the positive x-axis is $(1,0)$.
Given that,
$(2\cos \theta -1{)}^2+3{\sin }^2\theta =\frac{25}{16}$
$\Rightarrow \cos \theta =\frac{3}{4}$
Or $\Rightarrow \theta =co{s}^{-1}\frac{3}{4}$