A rod of length 6 units slides with its ends | vedclass

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(A) Let $A(a, 0)$ and $B(0, b)$ be the end points of the rod and $P(h, k)$ be the midpoint.Since $P$ is the midpoint,we have $h = \frac{a+0}{2} = \fra

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$x^2+y^2=9$

Vedclass · Answer

(A) Let $A(a, 0)$ and $B(0, b)$ be the end points of the rod and $P(h, k)$ be the midpoint.Since $P$ is the midpoint,we have $h = \frac{a+0}{2} = \frac{a}{2}$ and $k = \frac{0+b}{2} = \frac{b}{2}$.Thus,$a = 2h$ and $b = 2k$.The length of the rod is given as $6$ units,so $\sqrt{(a-0)^2 + (0-b)^2} = 6$.Squaring both sides,we get $a^2 + b^2 = 36$.Substituting the values of $a$ and $b$,we get $(2h)^2 + (2k)^2 = 36$.$4h^2 + 4k^2 = 36$.Dividing by $4$,we get $h^2 + k^2 = 9$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 9$.

$A$ rod of length $6$ units slides with its ends on the coordinate axes. The locus of the midpoint of the rod is

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