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A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

$\begin {array} {1 1} (A)\;x^2+y^2=729 & \quad (B)\;x^2=9y \\ (C)\;\large\frac{x^2}{81}+\large\frac{y^2}{9}=1 & \quad (D)\;\large\frac{x^2}{81}-\large\frac{y^2}{9}=1 \end {array}$

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  • Equation of an ellipse whose major axis is along the x - axis is given by $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
Step 1 :
Let AB be the rod making an angle $ \theta $ with x - axis, and P(x, y) be a point on it such that AP = 3 cm
Now PB = AB - AP is (AB = 12 cm )
$ \therefore $ PB = 12-3 cm
= 9cm
Draw perpendicular from PQ and PR such that $ PQ \perp 0Y$ and $PR \perp 0X$
Step 2 :
Consider the $ \Delta PBQ $
$ \cos \theta = \large\frac{PQ}{PB} = \large\frac{x}{9}$
Consider the $ \Delta PRA$
$ \sin \theta = \large\frac{PR}{PA}$$=\large\frac{x}{3}$
Step 3 :
We know $\sin^2 \theta + \cos^2 \theta = 1$
Hence substituting this we get,
$ \bigg( \large\frac{y}{3} \bigg)^2 + \bigg( \large\frac{x}{9} \bigg)^2=1$
$ \Rightarrow \large\frac{x^2}{81}$$ + \large\frac{y^2}{9}$$=1$
Thus the equation of the locus of a point p on the rod is $ \large\frac{x^2}{81}$$+\large\frac{y^2}{9}$$=1$
answered Jul 23, 2014 by thanvigandhi_1

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