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If \mathrm{P} is \mathrm{(a \cos \theta, b \sin \theta)} and \mathrm{Q} is \mathrm{( a \cos \phi, b \sin \phi)}, and the chord \mathrm{P Q} intersects major axis of \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at a distance \mathrm{c} from the centre, then \mathrm{\tan \frac{\alpha}{2} \tan \frac{\beta}{2}-k \frac{c-a}{c+a}=0}, where \mathrm{k=}

 

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

The equation of the chord \mathrm{P Q} is

\mathrm{ \frac{x}{a} \cos \left(\frac{\theta+\phi}{2}\right)+\frac{y}{b} \sin \left(\frac{\theta+\phi}{2}\right)}

\mathrm{ =\cos \left(\frac{\theta-\phi}{2}\right)}           (standard equation)

The point of intersection with x-axis is \mathrm{M(c, 0)}. The condition is \mathrm{ \frac{c}{a} \cos \left(\frac{\theta+\phi}{2}\right)=\cos \left(\frac{\theta-\phi}{2}\right)}

\mathrm{ \Rightarrow \frac{\cos \frac{\theta+\phi}{2}}{\cos \frac{\theta-\phi}{2}}=\frac{a}{c} }

Applying componendo-dividendo, we obtain

\mathrm{ \frac{\cos \frac{\theta+\phi}{2}+\cos \frac{\theta-\phi}{2}}{\cos \frac{\theta+\phi}{2}-\cos \frac{\theta-\phi}{2}}=\frac{a+c}{a-c} }

\mathrm{ \text { Or } \quad \frac{2 \cos \frac{\theta}{2} \cos \frac{\phi}{2}}{-2 \sin \frac{\theta}{2} \sin \frac{\phi}{2}}=\frac{a+c}{a-c} }

\mathrm{ \Rightarrow \quad \tan \frac{\theta}{2} \tan \frac{\phi}{2}=\frac{c-a}{c+a} }

\mathrm{ \therefore \quad k=1 }

Hene option 1 is correct.




 



 

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Rakesh

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