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  • If the eccentric angles of the ends of a focal chord of the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

If the eccentric angles of the ends of a focal chord of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)}are \mathrm{\theta_1 \, \, and\, \, \theta_2}, then value of \mathrm{\tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}} equals

Option: 1

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


Option: 2

\mathrm{\frac{e-1}{e^2+1}}


Option: 3

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


Option: 4

\mathrm{\frac{e^2+1}{e-1}}


Answers (1)

best_answer

Equation of the chord joining the points

\mathrm{\left(a \cos \theta_1, b \sin \theta_1\right) \&\left(a \cos \theta_2, b \sin \theta_2\right)} is given by

\mathrm{ \frac{x}{a} \cos \left(\frac{\theta_1+\theta_2}{2}\right)+\frac{y}{b} \sin \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \left(\frac{\theta_1-\theta_2}{2}\right) \text {, } }

which passes through \mathrm{(a e, 0),} we have
 

\mathrm{\begin{gathered} e \cos \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \left(\frac{\theta_1-\theta_2}{2}\right) \\\\ \Rightarrow \quad e\left[\cos \frac{\theta_1}{2} \cos \frac{\theta_2}{2}-\sin \frac{\theta_1}{2} \sin \frac{\theta_2}{2}\right] \\\\ =\cos \frac{\theta_1}{2} \cos \frac{\theta_2}{2}+\sin \frac{\theta_1}{2} \sin \frac{\theta_2}{2} \end{gathered} }

 

\mathrm{\begin{aligned} & \Rightarrow \quad e\left[1-\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}\right]=1+\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} \\\\ & \Rightarrow \quad(e-1)=(e+1) \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} \\\\ & \Rightarrow \quad \tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}=\frac{e-1}{e+1} \end{aligned}}

Posted by

Ritika Jonwal

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