Question

If the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ are ${\theta }_1$ and ${\theta }_2$, then value of $\tan \frac{{\theta }_1}{2}\times \tan \frac{{\theta }_2}{2}$ equals to:

• A. $\frac{e-1}{e+1}$
• B. $\frac{e-1}{e^2+1}$
• C. $\frac{e+1}{e-1}$
  
• D. $\frac{e^2+1}{e-1}$

Answer 1

Answer: (A)

$\frac{e-1}{e+1}$

Equation of the chord joining the points $(a\cos {\theta }_1,b\sin {\theta }_1)$ and $(a\cos {\theta }_2,b\sin {\theta }_2)$ is given by

$\frac{x}{a}\cos \frac{{\theta }_1+{\theta }_2}{2}+\frac{y}{b}\sin \frac{{\theta }_1+{\theta }_2}{2}=\cos \frac{{\theta }_1-{\theta }_2}{2}$

which passes through $(ae,0)$, we have

$e\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\cos \frac{{\theta }_1-{\theta }_2}{2}$

$\Rightarrow e[\cos \frac{{\theta }_1}{2}\cos \frac{{\theta }_2}{2}-\sin \frac{{\theta }_1}{2}\sin \frac{{\theta }_2}{2}]=\cos \frac{{\theta }_1}{2}\cos \frac{{\theta }_2}{2}+\sin \frac{{\theta }_1}{2}\sin \frac{{\theta }_2}{2}$

$\Rightarrow e[1-\tan \frac{{\theta }_1}{2}\tan \frac{{\theta }_2}{2}]=1+\tan \frac{{\theta }_1}{2}\tan \frac{{\theta }_2}{2}$

$\Rightarrow (e-1)=(e+1)\tan \frac{{\theta }_1}{2}\tan \frac{{\theta }_2}{2}\Rightarrow \tan \frac{{\theta }_1}{2}\tan \frac{{\theta }_2}{2}=\frac{e-1}{e+1}$