The line joining \( \Large A \left(b \cos \alpha ,\ b \sin \alpha \right)\ and\ B \left(a \cos \beta ,\ a \sin \beta \right) \) is produced to the point \( \Large M \left(x,\ y\right) \) so that \( \Large AM:MB=b:a\ then\ x \cos \left(\frac{ \alpha + \beta }{2}\right)\ +\ y \sin \left(\frac{ \alpha + \beta }{2}\right) \) is:
Correct Answer: Description for Correct answer:
Therefore, M divides AB in the ratio b : a (externally)
\( \Large x=\frac{ba \cos \beta - ab \cos \alpha }{b-a}\ and\ y=\frac{ab \sin \beta - ab \sin \alpha }{b-a} \)
\( \Large \frac{x}{y} = \frac{\cos \beta - \cos \alpha }{\sin \beta - \sin \alpha } \)
=> \( \Large \frac{x}{y}=\frac{2\sin \left(\frac{ \alpha + \beta }{2}\right)\sin \left(\frac{ \alpha - \beta }{2}\right) }{2\cos \left(\frac{\alpha + \beta}{2} \right)sin \left(\frac{ \beta - \alpha }{2}\right) } \)
=> \( \Large x \cos \frac{ \alpha + \beta }{2}+y \sin \left(\frac{ \beta - \alpha }{2}\right)=0 \)
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