A(a, 0) is a fixed point and theta is a param | vedclass

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(A) Let the centroid of $	riangle APQ$ be $(h, k)$.Given vertices are $A(a, 0)$,$P(a \cos 	heta, a \sin 	heta)$,and $Q(b \sin 	heta, -b \cos 	het

Vedclass · Accepted Answer

a circle with centre at $\left(\frac{a}{3}, 0\right)$ and radius $\frac{\sqrt{a^2+b^2}}{3}$

Vedclass · Answer

(A) Let the centroid of $	riangle APQ$ be $(h, k)$.Given vertices are $A(a, 0)$,$P(a \cos 	heta, a \sin 	heta)$,and $Q(b \sin 	heta, -b \cos 	heta)$.The centroid $(h, k)$ is given by:$h = \frac{a + a \cos 	heta + b \sin 	heta}{3} \implies 3h - a = a \cos 	heta + b \sin 	heta$$k = \frac{0 + a \sin 	heta - b \cos 	heta}{3} \implies 3k = a \sin 	heta - b \cos 	heta$Squaring and adding the two equations:$(3h - a)^2 + (3k)^2 = (a \cos 	heta + b \sin 	heta)^2 + (a \sin 	heta - b \cos 	heta)^2$$(3h - a)^2 + 9k^2 = a^2(\cos^2 	heta + \sin^2 	heta) + b^2(\sin^2 	heta + \cos^2 	heta) + 2ab \cos 	heta \sin 	heta - 2ab \sin 	heta \cos 	heta$$(3h - a)^2 + 9k^2 = a^2 + b^2$$9(h - \frac{a}{3})^2 + 9k^2 = a^2 + b^2$$(h - \frac{a}{3})^2 + k^2 = \frac{a^2 + b^2}{9} = \left(\frac{\sqrt{a^2 + b^2}}{3}ight)^2$Thus,the locus is a circle with centre at $\left(\frac{a}{3}, 0ight)$ and radius $\frac{\sqrt{a^2 + b^2}}{3}$.

$A(a, 0)$ is a fixed point and $\theta$ is a parameter such that $0 < \theta < 2 \pi$. If $P(a \cos \theta, a \sin \theta)$ is a point on the circle $x^2+y^2=a^2$ and $Q(b \sin \theta, -b \cos \theta)$ is a point on the circle $x^2+y^2=b^2$,then the locus of the centroid of the triangle $APQ$ is

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