$A$ rod $AB$ of length $15\, cm$ rests in between two coordinate axes in such a way that the end point $A$ lies on $x$-axis and end point $B$ lies on $y$-axis. A point $P(x, y)$ is taken on the rod in such a way that $AP = 6\, cm$. Find the locus of $P$ of an ellipse.
Updated On: Jul 6, 2022
$\frac{x^{2}}{9}+\frac{y^{2}}{6}=1$
$\frac{x^{2}}{36}+\frac{y^{2}}{81}=1$
$\frac{x^{2}}{81}+\frac{y^{2}}{36}=1$
$\frac{y^{2}}{9}+\frac{x^{2}}{6}=1$
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The Correct Option isC
Solution and Explanation
Let $AB$ be the rod making an angle $\theta$ with $OX$ as shown in figure and $P(x, y)$ is the point on it such that $AP = 6 \,cm$.
Since $AB = 15\, cm$, we have $PB = 9\, cm$
From $P$ draw $PQ$ and $PR$ perpendicular on $y$-axis and $x$-axis, respectively.
From $\Delta PBQ$, $cos\theta=\frac{x}{9}$
From $\Delta PRA$, $sin\theta=\frac{y}{6}$
Since $cos^{2}\theta+sin^{2}\theta=1$$\Rightarrow\, \left(\frac{x}{9}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1$
or $ \frac{x^{2}}{81}+\frac{y^{2}}{36}=1$
