The line joining A(b cos alpha, b sin alpha) | vedclass

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(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$,the coord

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(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$,the coordinates of $M$ are given by the section formula for external division:$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$Now,consider the expression $E = x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$.Substituting $x$ and $y$:$E = \frac{ab}{b-a} [(\cos \beta - \cos \alpha) \cos \frac{\alpha+\beta}{2} + (\sin \beta - \sin \alpha) \sin \frac{\alpha+\beta}{2}]$Using the identity $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$E = \frac{ab}{b-a} [(-2 \sin \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \cos \frac{\alpha+\beta}{2} + (2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \sin \frac{\alpha+\beta}{2}]$$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [-\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} + \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha+\beta}{2}]$$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [0] = 0$.

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 $M(x, y)$ such that $AM : MB = b : a$. Then,$x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$ is equal to

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