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VITEEE 2010 Paper

Question : 96 of 120

Marks: +1, -0

l

Solution:

Let both of the ends of the rod are on x - axis and y - axis. Let AB be rod of length

l

and coordinates of A and B be (a,0) and (0,b) respectively.

Let P(h,k) be the mid point of the rod AB.

Then,

\left[ \begin{array}{l} h = \frac{0+a}{2} = \frac{a}{2} \\ k = \frac{b+0}{2} = \frac{b}{2} \end{array} \right] \ldots (i)

In Δ OAB,

OA^{2}+OB^{2}=AB^{2}

⇒

a^{2}+b^{2}=l^{2}

⇒

(2h)^{2}+(2k)^{2}=l^{2}\;\;\;\;[\text{using eq.}(i)]

⇒

h^{2}+k^{2}=\frac{l^{2}}{4}

∴ The equation of locus is

x^{2}+y^{2}=\frac{l^{2}}{4}

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