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The sum of the eccentric angles of the conormal points on the ellipse $\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ is equal to

Option: 1
$\mathrm{2 n \pi, n \in N}$

Option: 2
$\mathrm{(2 n+1) \pi, n \in N}$

Option: 3
$\mathrm{n \pi, n \in N}$

Option: 4
$\frac{n}{2} \pi, n \in N$

Answers (1)

best_answer

As obtained in the last solved example (8), the normal at $\mathrm{P(a \cos \theta, b \sin \theta)}$ passes through $\mathrm{M\left(x_1, y_1\right)}$ . The condition is

$\mathrm{ b y_1 t^4+2 t^3\left(a x_1+a^2 e^2\right)+2 t\left(a x_1-a^2 e^2\right)-b y_1=0 }$

The roots are $\mathrm{ t_1, t_2, t_3, t_4 }$

$\mathrm{ \sum t_1 =-\frac{2\left(a x_1+a^2 e^2\right)}{b y_1} }$

$\mathrm{ \sum t_1 t_2 =0 }$ ........(1)

$\mathrm{ \sum t_1 t_2 t_3 =-\frac{2\left(a x_1-a^2 e^2\right)}{b y_1} }$

$\mathrm{ t_1 t_2 t_3 t_4 =-1 }$ ...........(2)

Let $\mathrm{ t_r=\tan \frac{\theta_r}{2}(r=1,2,3, r) }$

$\mathrm{ \therefore \quad \tan \left(\frac{\theta_1}{2}+\frac{\theta_2}{2}+\frac{\theta_3}{2}+\frac{\theta_4}{2}\right) }$

$\mathrm{ =\frac{\sum \tan \frac{\theta_1}{2}-\sum \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} \tan \frac{\theta_3}{2}}{1-\sum \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}+\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} \tan \frac{\theta_3}{2} \tan \frac{\theta_4}{2}} }$

$\mathrm{ =\frac{\sum t_1-\sum t_1 t_2 t_3}{1-\sum t_1 t_2+t_1 t_2 t_3 t_4} }$

$\mathrm{ =\frac{\sum t_1-\sum t_1 t_2 t_3}{1-0+(-1)}=\infty \text { from eqs. (1) and (2) } }$

$\mathrm{ \Rightarrow \quad \frac{\theta_1+\theta_2+\theta_3+\theta_4}{2}=\frac{\pi}{2}+n \pi }$

$\mathrm{ \text { Or } \quad \theta_1+\theta_2+\theta_3+\theta_4 =2 n \pi+\pi }$

$\mathrm{ =(2 n+1) \pi, n \in N }$

This is a standard property about the eccentric angles for normals of an ellipse.

Hence option 2 is correct.

Posted by

Pankaj

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