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The line joining $A(b \cos \alpha, b \sin \alpha)$
and
$B(a \cos \beta, a \sin \beta),$ where $a \neq b,$ is produced to the
point $M(x, y)$ so that $A M: M B=b: a$. Then, $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}$ is equal to
Options:
and
$B(a \cos \beta, a \sin \beta),$ where $a \neq b,$ is produced to the
point $M(x, y)$ so that $A M: M B=b: a$. Then, $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}$ is equal to
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Since, $A M: B M=b: a$ $\therefore M$ divides $A B$ extemally in the ratio $b: a$...(i)
$\therefore \quad x=\frac{b \cdot a \cos \beta-a b \cos \alpha}{b-a}$
$y=\frac{b a \sin \beta-a b \sin \alpha}{b-a}$
(ii)
Divide Eq. (i) by Eq, (ii), we get
$\frac{x}{y}=\frac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}=\frac{2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{-2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}$
$\therefore \quad x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$
$\therefore \quad x=\frac{b \cdot a \cos \beta-a b \cos \alpha}{b-a}$
$y=\frac{b a \sin \beta-a b \sin \alpha}{b-a}$
(ii)
Divide Eq. (i) by Eq, (ii), we get
$\frac{x}{y}=\frac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}=\frac{2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{-2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}$
$\therefore \quad x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$
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