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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Given $AB = 15$,so $\sqrt{a^2 + b^2} = 15$,which implies $a^2 + b^2 = 225$.Let $P(x, y)

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$x=9 \cos 	heta, y=6 \sin 	heta$

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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Given $AB = 15$,so $\sqrt{a^2 + b^2} = 15$,which implies $a^2 + b^2 = 225$.Let $P(x, y)$ be a point on $AB$ such that $\frac{AP}{PB} = \frac{2}{3}$.Using the section formula,$x = \frac{2(0) + 3(a)}{2+3} = \frac{3a}{5}$ and $y = \frac{2(b) + 3(0)}{2+3} = \frac{2b}{5}$.From these,$a = \frac{5x}{3}$ and $b = \frac{5y}{2}$.Substituting these into $a^2 + b^2 = 225$:$(\frac{5x}{3})^2 + (\frac{5y}{2})^2 = 225$.$\frac{25x^2}{9} + \frac{25y^2}{4} = 225$.Dividing by $25$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 9$,or $\frac{x^2}{81} + \frac{y^2}{36} = 1$.This represents an ellipse with $x = 9 \cos 	heta$ and $y = 6 \sin 	heta$.

$A$ line segment joining a point $A$ on $x$-axis to a point $B$ on $y$-axis is such that $AB=15$. If $P$ is a point on $AB$ such that $\frac{AP}{PB}=\frac{2}{3}$,then the locus of $P$ is:

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