If the line joining the points A(b cos alpha, | vedclass

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(C) The point $N(x, y)$ divides the line segment $AB$ externally in the ratio $b: a$.Using the external section formula,the coordinates of $N$ are giv

Vedclass · Accepted Answer

$x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$

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(C) The point $N(x, y)$ divides the line segment $AB$ externally in the ratio $b: a$.Using the external section formula,the coordinates of $N$ are given by:$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$From these,we have:$\frac{x}{ab} = \frac{\cos \beta - \cos \alpha}{b - a} \implies \frac{b - a}{ab} = \frac{\cos \beta - \cos \alpha}{x}$$\frac{y}{ab} = \frac{\sin \beta - \sin \alpha}{b - a} \implies \frac{b - a}{ab} = \frac{\sin \beta - \sin \alpha}{y}$Equating the two expressions for $\frac{b - a}{ab}$:$\frac{\cos \beta - \cos \alpha}{x} = \frac{\sin \beta - \sin \alpha}{y}$$y(\cos \beta - \cos \alpha) = x(\sin \beta - \sin \alpha)$Using the sum-to-product formulas:$y \left( -2 \sin \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} ight) = x \left( 2 \cos \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} ight)$Assuming $\sin \frac{\beta - \alpha}{2} 
eq 0$,we divide both sides by $2 \sin \frac{\beta - \alpha}{2}$:$-y \sin \frac{\alpha + \beta}{2} = x \cos \frac{\alpha + \beta}{2}$$x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = 0$

If the line joining the points $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ is extended to the point $N(x, y)$ such that $AN: NB = b: a$,then

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