Question

If $\frac{\pi }{3},\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$, then $\tan \theta =$

• A. $-\sqrt{3}$
• B. $\sqrt{3}$
• C. $-1$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$
Let eccentricity of ellipse be ' $e^{\prime }$
Then $b^2=a^2\left(1-e^2\right)$
Here, $b^2=12,a^2=16$
$\therefore 12=16\left(1-e^2\right)$
$1-e^2=\frac{3}{4}$
or $e^2=1-\frac{3}{4}=\frac{1}{4}$
or $e=\frac{1}{2}$
If $\alpha ,\beta$ are the eccentric angles of the ends of a focal chord of the ellipse, then eccentricity is given by
$e=\frac{\cos \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)}$
Here, $\alpha =\frac{\pi }{3},\beta =\theta$, and $e=\frac{1}{2}$
$\frac{1}{2}=\frac{\cos \left(\frac{\pi /3-\theta }{2}\right)}{\cos \left(\frac{\pi /3+\theta }{2}\right)}$
Multiplying in $N^r$ and $D^r$ by $2\sin \left(\frac{\pi /3+\theta }{2}\right)$
$\frac{1}{2}=\frac{2\sin \left(\frac{\pi /3+\theta }{2}\right)\cos \left(\frac{\pi /3-\theta }{2}\right)}{2\sin \left(\frac{\pi /3+\theta }{2}\right)\cos \left(\frac{\pi /3+\theta }{2}\right)}$
$=\frac{\sin \left(\frac{\pi /3+\theta +\pi /3-\theta }{2}\right)+\sin \left(\frac{\pi /3+\theta -\pi /3+\theta }{2}\right)}{\sin (\pi /3+\theta )}$
$\{\because 2\sin A\cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)$
and $2\sin A\cos A=\sin 2A\}$
$\frac{1}{2}=\frac{\sin \frac{\pi }{3}+\sin \theta }{\sin \frac{\pi }{3}\cos \theta +\cos \frac{\pi }{3}\sin \theta }$
$\sin \frac{\pi }{3}\cos \theta +\cos \frac{\pi }{3}\sin \theta =2\left(\sin \frac{\pi }{3}+\sin \theta \right)$
$\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta =2\left(\frac{\sqrt{3}}{2}+\sin \theta \right)$
$\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta =\sqrt{3}+2\sin \theta$
or $\frac{\sqrt{3}}{2}\cos \theta =\frac{3}{2}\sin \theta +\sqrt{3}$
$\cos \theta =\sqrt{3}\sin \theta +2$ or $\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta =1$
or $\cos \frac{\pi }{3}\cos \theta -\sin \frac{\pi }{3}\sin \theta =1$
or $\cos \left(\frac{\pi }{3}+\theta \right)=1$ or $\cos \left(\frac{\pi }{3}+\theta \right)=\cos 0^{\circ }$
On comparing both sides, we get
$\frac{\pi }{3}+\theta =0^{\circ }$ or $\theta =-\frac{\pi }{3}\therefore \tan \theta =\tan \left(\frac{-\pi }{3}\right)$
$\tan \theta =-\sqrt{3}$