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(B) Let a point on the circle be $P(\cos 	heta, \sin 	heta, 0)$.Let the foot of the perpendicular from $P$ to the plane $2x + 3y + z = 6$ be $Q(h, k

Vedclass · Accepted Answer

$(5x + 6y - 12)^{2} + 4(3x + 5y - 9)^{2} = 1, z = 6 - 2x - 3y$

Vedclass · Answer

(B) Let a point on the circle be $P(\cos 	heta, \sin 	heta, 0)$.Let the foot of the perpendicular from $P$ to the plane $2x + 3y + z = 6$ be $Q(h, k, w)$.The line $PQ$ is perpendicular to the plane,so its direction ratios are proportional to the normal of the plane $(2, 3, 1)$.Thus,$\frac{h - \cos 	heta}{2} = \frac{k - \sin 	heta}{3} = \frac{w - 0}{1} = \lambda$.Since $Q(h, k, w)$ lies on the plane $2x + 3y + z = 6$,we have $2h + 3k + w = 6$.Substituting $h = \cos 	heta + 2\lambda$,$k = \sin 	heta + 3\lambda$,and $w = \lambda$ into the plane equation:$2(\cos 	heta + 2\lambda) + 3(\sin 	heta + 3\lambda) + \lambda = 6$$2\cos 	heta + 3\sin 	heta + 14\lambda = 6 \implies \lambda = \frac{6 - 2\cos 	heta - 3\sin 	heta}{14}$.Then $h = \cos 	heta + 2\left(\frac{6 - 2\cos 	heta - 3\sin 	heta}{14}ight) = \frac{14\cos 	heta + 12 - 4\cos 	heta - 6\sin 	heta}{14} = \frac{10\cos 	heta - 6\sin 	heta + 12}{14}$.$k = \sin 	heta + 3\left(\frac{6 - 2\cos 	heta - 3\sin 	heta}{14}ight) = \frac{14\sin 	heta + 18 - 6\cos 	heta - 9\sin 	heta}{14} = \frac{5\sin 	heta - 6\cos 	heta + 18}{14}$.Rearranging these gives $5h + 6k = \frac{50\cos 	heta - 30\sin 	heta + 60 + 30\sin 	heta - 36\cos 	heta + 108}{14} = \frac{14\cos 	heta + 168}{14} = \cos 	heta + 12 \implies \cos 	heta = 5h + 6k - 12$.Similarly,$3h + 5k = \frac{30\cos 	heta - 18\sin 	heta + 36 + 25\sin 	heta - 30\cos 	heta + 90}{14} = \frac{7\sin 	heta + 126}{14} = \frac{\sin 	heta}{2} + 9 \implies \sin 	heta = 2(3h + 5k - 9)$.Using $\cos^{2} 	heta + \sin^{2} 	heta = 1$,we get $(5h + 6k - 12)^{2} + 4(3h + 5k - 9)^{2} = 1$.

The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

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