If the chord through the point whose eccentri | vedclass

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Question

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

Vedclass · Accepted Answer

$e - 1$

Vedclass · Answer

Equation of chord to the ellipse with eccentric angle $	heta$ and $\phi$ is given by, $\frac{x}{a} \cos \frac{	heta+\phi}{2}+\frac{y}{b} \sin \frac{	heta+\phi}{2}=\cos \frac{	heta-\phi}{2}$

Given it passes through $(a e, 0)$ $\Rightarrow e \cos \frac{	heta+\phi}{2}=\cos \frac{	heta-\phi}{2}$

$\Rightarrow e\left(\cos \frac{	heta}{2} \cos \frac{\phi}{2}-\sin \frac{	heta}{2} \sin \frac{\phi}{2}ight)=\cos \frac{	heta}{2} \cos \frac{\phi}{2}+\sin \frac{	heta}{2} \sin \frac{\phi}{2}$

$\Rightarrow e\left(1-	an \frac{	heta}{2} 	an \frac{\phi}{2}ight)=1+	an \frac{	heta}{2} 	an \frac{\phi}{2}$

$	herefore 	an \frac{	heta}{2} 	an \frac{\phi}{2}=\frac{e-1}{e+1}$

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

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