If the chord through the point whose eccentric angles are , & , o… - Vidyadip

MCQ

If the chord through the point whose eccentric angles are $\theta \,\& \,\phi $ on the ellipse,$(x^2/a^2) + (y^2/b^2) = 1$ passes through the focus, then the value of $ (1 + e)$ $\tan(\theta /2) \tan(\phi /2)$ is

A
$e + 1$
✓
$e - 1$
C
$1 - e$
D
$0$

✓

Answer

Correct option: B.

$e - 1$

b
Equation of chord to the ellipse with eccentric angle $\theta$ and $\phi$ is given by, $\frac{x}{a} \cos \frac{\theta+\phi}{2}+\frac{y}{b} \sin \frac{\theta+\phi}{2}=\cos \frac{\theta-\phi}{2}$

Given it passes through $(a e, 0)$ $\Rightarrow e \cos \frac{\theta+\phi}{2}=\cos \frac{\theta-\phi}{2}$

$\Rightarrow e\left(\cos \frac{\theta}{2} \cos \frac{\phi}{2}-\sin \frac{\theta}{2} \sin \frac{\phi}{2}\right)=\cos \frac{\theta}{2} \cos \frac{\phi}{2}+\sin \frac{\theta}{2} \sin \frac{\phi}{2}$

$\Rightarrow e\left(1-\tan \frac{\theta}{2} \tan \frac{\phi}{2}\right)=1+\tan \frac{\theta}{2} \tan \frac{\phi}{2}$

$\therefore \tan \frac{\theta}{2} \tan \frac{\phi}{2}=\frac{e-1}{e+1}$

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