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(B) Let $AB$ be the rod of length $12 \, cm$. Let $A$ be on the $x-$axis and $B$ be on the $y-$axis. Let $P(x, y)$ be a point on the rod such that $AP

Vedclass · Accepted Answer

$\frac{x^{2}}{9} + \frac{y^{2}}{81} = 1$

Vedclass · Answer

(B) Let $AB$ be the rod of length $12 \, cm$. Let $A$ be on the $x-$axis and $B$ be on the $y-$axis. Let $P(x, y)$ be a point on the rod such that $AP = 3 \, cm$. Then $PB = AB - AP = 12 - 3 = 9 \, cm$.Let $\angle OAB = 	heta$. Then in $	riangle PRA$ (where $PR \perp OA$),we have $\cos 	heta = \frac{AR}{AP} = \frac{x}{3}$,so $x = 3 \cos 	heta$.In $	riangle PQB$ (where $PQ \perp OB$),we have $\sin 	heta = \frac{QB}{PB} = \frac{y}{9}$,so $y = 9 \sin 	heta$.Using the identity $\cos^{2} 	heta + \sin^{2} 	heta = 1$,we get $\left(\frac{x}{3}ight)^{2} + \left(\frac{y}{9}ight)^{2} = 1$.Thus,the equation of the locus is $\frac{x^{2}}{9} + \frac{y^{2}}{81} = 1$.

$A$ rod of length $12 \, cm$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod,which is $3 \, cm$ from the end in contact with the $x-$axis.

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