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प्रश्न
A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.
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उत्तर
Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.
As the rod AB slides, the values of a and b change. Let P(h, k) be a point on AB.
Here, BP:AP = 1:2 .
\[\therefore h = \frac{a + 0}{3}, k = \frac{0 + 2b}{3}\]
\[ \Rightarrow a = 3h, b = \frac{3k}{2}\] ... (1)
The length of the given rod is l.
\[\therefore AB = l\]
\[ \Rightarrow \sqrt{a^2 + b^2} = l\]
\[ \Rightarrow a^2 + b^2 = l^2\]
Using equation (1), we get:
\[\Rightarrow 9 h^2 + \left( \frac{3k}{2} \right)^2 = l^2 \]
\[ \Rightarrow h^2 + \frac{k^2}{4} = \frac{l^2}{9}\]
Hence, the locus of (h, k) is \[x^2 + \frac{y^2}{4} = \frac{l^2}{9}\]
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