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(B) Let point $A(a, 0)$ be on the $x$-axis and $B(0, b)$ be on the $y$-axis.Let $P(h, k)$ divide $AB$ in the ratio $1 : 2$.By the section formula,we h

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an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

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(B) Let point $A(a, 0)$ be on the $x$-axis and $B(0, b)$ be on the $y$-axis.Let $P(h, k)$ divide $AB$ in the ratio $1 : 2$.By the section formula,we have:$h = \frac{2(0) + 1(a)}{1 + 2} = \frac{a}{3} \Rightarrow a = 3h$$k = \frac{2(b) + 1(0)}{1 + 2} = \frac{2b}{3} \Rightarrow b = \frac{3k}{2}$Since the length of the staircase is $l$,we have $a^2 + b^2 = l^2$.Substituting the values of $a$ and $b$:$(3h)^2 + (\frac{3k}{2})^2 = l^2$$9h^2 + \frac{9k^2}{4} = l^2$$\frac{h^2}{(l/3)^2} + \frac{k^2}{(2l/3)^2} = 1$This is the equation of an ellipse with semi-major axis $b' = \frac{2l}{3}$ and semi-minor axis $a' = \frac{l}{3}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{(a')^2}{(b')^2}} = \sqrt{1 - \frac{(l/3)^2}{(2l/3)^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

$A$ staircase of length $l$ rests against a vertical wall and a floor of a room. Let $P$ be a point on the staircase,nearer to its end on the wall,that divides its length in the ratio $1 : 2$. If the staircase begins to slide on the floor,then the locus of $P$ is

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