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(D) Let $A(a, 0)$ and $B(0, b)$ be the points on the axes,and let $P(h, k)$ be the point that divides the line segment $AB$ in the ratio $2:1$.Using t

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$4x^2+y^2=16$

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(D) Let $A(a, 0)$ and $B(0, b)$ be the points on the axes,and let $P(h, k)$ be the point that divides the line segment $AB$ in the ratio $2:1$.Using the section formula,the coordinates of $P$ are given by:$P(h, k) = \left(\frac{1 \cdot a + 2 \cdot 0}{2+1}, \frac{1 \cdot 0 + 2 \cdot b}{2+1}\right) = \left(\frac{a}{3}, \frac{2b}{3}\right)$Equating the coordinates,we get:$h = \frac{a}{3} \Rightarrow a = 3h$$k = \frac{2b}{3} \Rightarrow b = \frac{3k}{2}$Given that the length of the line segment $AB = 6$,we have:$\sqrt{(a-0)^2 + (0-b)^2} = 6$$a^2 + b^2 = 36$Substituting the values of $a$ and $b$ in terms of $h$ and $k$:$(3h)^2 + \left(\frac{3k}{2}\right)^2 = 36$$9h^2 + \frac{9k^2}{4} = 36$Dividing by $9$:$h^2 + \frac{k^2}{4} = 4$$4h^2 + k^2 = 16$Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $4x^2 + y^2 = 16$.

$A$ straight line meets the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If $AB = 6$ units,then the locus of the point $P$ which divides the line segment $AB$ such that $AP : PB = 2 : 1$ is

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