P(theta_1) and Q(theta_2) are two points on t | vedclass

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(C) For a focal chord $PSQ$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passing through the focus $S(ae, 0)$,the relation between the eccentric

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}},\left(\frac{a}{\sqrt{2}}, 0\right)$

Vedclass · Answer

(C) For a focal chord $PSQ$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passing through the focus $S(ae, 0)$,the relation between the eccentric angles $	heta_1$ and $	heta_2$ is given by $	an \left(\frac{	heta_1}{2}ight) 	an \left(\frac{	heta_2}{2}ight) = \frac{e-1}{e+1}$.Given $	an \left(\frac{	heta_1}{2}ight) 	an \left(\frac{	heta_2}{2}ight) = -(2\sqrt{2}+3)$.So,$\frac{e-1}{e+1} = -(2\sqrt{2}+3)$.Let $k = 2\sqrt{2}+3$. Then $\frac{e-1}{e+1} = -k$.$e-1 = -ke - k$ $\Rightarrow e(1+k) = 1-k$ $\Rightarrow e = \frac{1-k}{1+k}$.$e = \frac{1-(2\sqrt{2}+3)}{1+(2\sqrt{2}+3)} = \frac{-2-2\sqrt{2}}{4+2\sqrt{2}} = \frac{-2(1+\sqrt{2})}{2(2+\sqrt{2})} = \frac{-(1+\sqrt{2})}{\sqrt{2}(\sqrt{2}+1)} = -\frac{1}{\sqrt{2}}$.Since eccentricity $e$ must be positive,we take the magnitude,so $e = \frac{1}{\sqrt{2}}$.The focus $S$ is $(ae, 0)$ or $(-ae, 0)$.Substituting $e = \frac{1}{\sqrt{2}}$,we get $S = \left(\pm \frac{a}{\sqrt{2}}, 0ight)$.Comparing with the options,option $C$ is $\left(\frac{a}{\sqrt{2}}, 0ight)$ and option $D$ is $\left(\frac{-a}{\sqrt{2}}, 0ight)$. Given the standard form,both are valid,but usually,the positive focus is considered. Based on the provided options,$C$ is the correct choice.

$P(\theta_1)$ and $Q(\theta_2)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e$. If $PSQ$ is a focal chord and $\tan \left(\frac{\theta_1}{2}\right) \tan \left(\frac{\theta_2}{2}\right)=-(2 \sqrt{2}+3)$,then $e$ and $S$ are

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