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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 $\tan \theta=$
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The correct answer is:
$-\sqrt{3}$
Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$
Let eccentricity of ellipse be ' $e$ ' Then
$b^2=a^2\left(1-e^2\right)$
Here, $\quad b^2=12, a^2=16$
$\therefore \quad 12=16\left(1-e^2\right)$
$1-e^2=\frac{3}{4}$
or $e^2=1-\frac{3}{4}=\frac{1}{4} \text { or } e=\frac{1}{2}$
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 $2 \sin \left(\frac{\pi / 3+\theta}{2}\right)$
$\frac{1}{2}=\frac{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3-\theta}{2}\right)}{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3+\theta}{2}\right)}$
$=\frac{\sin \left(\frac{\pi / 3+\theta+\pi / 3-\theta}{2}\right)+\sin \left(\frac{\pi / 3+\theta-\pi / 3+\theta}{2}\right)}{\sin (\pi / 3+\theta)}$
$\left\{\because 2 \sin A \cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right.$ and $2 \sin A \cos A=\sin 2 A\}$
$\frac{1}{2}=\frac{\sin \frac{\pi}{3}+\sin \theta}{\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta}$
$\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=2\left(\sin \frac{\pi}{3}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=2\left(\frac{\sqrt{3}}{2}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\sqrt{3}+2 \sin \theta$
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 $\quad \cos \frac{\pi}{3} \cos \theta-\sin \frac{\pi}{3} \sin \theta=1$
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
$\frac{\pi}{3}+\theta=0^{\circ} \text { or } \theta=-\frac{\pi}{3} \quad \therefore \quad \tan \theta=\tan \left(\frac{-\pi}{3}\right)$
$\tan \theta=-\sqrt{3}$
Let eccentricity of ellipse be ' $e$ ' Then
$b^2=a^2\left(1-e^2\right)$
Here, $\quad b^2=12, a^2=16$
$\therefore \quad 12=16\left(1-e^2\right)$
$1-e^2=\frac{3}{4}$
or $e^2=1-\frac{3}{4}=\frac{1}{4} \text { or } e=\frac{1}{2}$
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 $2 \sin \left(\frac{\pi / 3+\theta}{2}\right)$
$\frac{1}{2}=\frac{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3-\theta}{2}\right)}{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3+\theta}{2}\right)}$
$=\frac{\sin \left(\frac{\pi / 3+\theta+\pi / 3-\theta}{2}\right)+\sin \left(\frac{\pi / 3+\theta-\pi / 3+\theta}{2}\right)}{\sin (\pi / 3+\theta)}$
$\left\{\because 2 \sin A \cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right.$ and $2 \sin A \cos A=\sin 2 A\}$
$\frac{1}{2}=\frac{\sin \frac{\pi}{3}+\sin \theta}{\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta}$
$\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=2\left(\sin \frac{\pi}{3}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=2\left(\frac{\sqrt{3}}{2}+\sin \theta\right)$
$\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\sqrt{3}+2 \sin \theta$
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 $\quad \cos \frac{\pi}{3} \cos \theta-\sin \frac{\pi}{3} \sin \theta=1$
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
$\frac{\pi}{3}+\theta=0^{\circ} \text { or } \theta=-\frac{\pi}{3} \quad \therefore \quad \tan \theta=\tan \left(\frac{-\pi}{3}\right)$
$\tan \theta=-\sqrt{3}$
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