The ends of a rod of length l move on two mut | vedclass

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(C) Let the rod be $AB$ with length $l$. Let the ends $A$ and $B$ lie on the $x$-axis and $y$-axis respectively,such that $A = (a, 0)$ and $B = (0, b)

Vedclass · Accepted Answer

$9{x^2} + 36{y^2} = 4{l^2}$

Vedclass · Answer

(C) Let the rod be $AB$ with length $l$. Let the ends $A$ and $B$ lie on the $x$-axis and $y$-axis respectively,such that $A = (a, 0)$ and $B = (0, b)$.Since the length of the rod is $l$,we have $a^2 + b^2 = l^2$.Let $P(h, k)$ be the point on the rod that divides it in the ratio $1 : 2$. Using the section formula,we have:$h = \frac{1 	imes 0 + 2 	imes a}{1 + 2} = \frac{2a}{3} \Rightarrow a = \frac{3h}{2}$$k = \frac{1 	imes b + 2 	imes 0}{1 + 2} = \frac{b}{3} \Rightarrow b = 3k$Substituting these into the equation $a^2 + b^2 = l^2$,we get:$(\frac{3h}{2})^2 + (3k)^2 = l^2$$\frac{9h^2}{4} + 9k^2 = l^2$$9h^2 + 36k^2 = 4l^2$Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 36y^2 = 4l^2$.

The ends of a rod of length $l$ move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio $1 : 2$ is

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