The coordinates of the foot of the perpendicu | vedclass

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(B) The line passes through $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$.Its equation is $x \cos \frac{\alpha + \beta}{2} + y \

Vedclass · Accepted Answer

$\left[ \frac{a}{2}(\cos \alpha + \cos \beta), \frac{a}{2}(\sin \alpha + \sin \beta) \right]$

Vedclass · Answer

(B) The line passes through $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$.Its equation is $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = a \cos \frac{\alpha - \beta}{2}$.The foot of the perpendicular from the origin $(0,0)$ to the line $x \cos 	heta + y \sin 	heta = p$ is $(p \cos 	heta, p \sin 	heta)$.Here,$p = a \cos \frac{\alpha - \beta}{2}$ and $	heta = \frac{\alpha + \beta}{2}$.Thus,the foot of the perpendicular is $\left( a \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}, a \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2} ight)$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ and $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we get:$x = \frac{a}{2} [\cos \alpha + \cos \beta]$ and $y = \frac{a}{2} [\sin \alpha + \sin \beta]$.Therefore,the correct option is $B$.

The coordinates of the foot of the perpendicular drawn from $(0,0)$ to the line joining $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ are

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