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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.

Posted by

Rakesh

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