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

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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$

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