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Consider a rigid square $\mathrm{ABCD}$ as in the figure with $\mathrm{A}$ and $\mathrm{B}$ on the $\mathrm{x}$ and y axis respectively.
When A and B slide along their respective axes, the locus of $\mathrm{C}$ forms a part of
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When A and B slide along their respective axes, the locus of $\mathrm{C}$ forms a part of
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Verified Answer
The correct answer is:
an ellipse which is not a circle
$\mathrm{C}(\mathrm{h}, \mathrm{k})$
$\mathrm{h}=\mathrm{a} \sin \theta \Rightarrow \sin \theta=\frac{\mathrm{h}}{\mathrm{a}}......(1)$
$\mathrm{k}=\mathrm{a} \sin \theta+\mathrm{a} \cos \theta$
$\mathrm{k}=\mathrm{h}+\mathrm{a} \cos \theta$
$\frac{\mathrm{k}-\mathrm{h}}{\mathrm{a}}=\cos \theta ......(2)$
$\begin{array}{l}
\sin ^{2} \theta+\cos ^{2} \theta=1 \\
\frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{(\mathrm{k}-\mathrm{h})^{2}}{\mathrm{a}^{2}}=1 \\
\mathrm{~h}^{2}+(\mathrm{k}-\mathrm{h})^{2}=\mathrm{a}^{2}
\end{array}$
locus $x^{2}+(y-x)^{2}=a^{2}$
solving we get $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{xy}=\mathrm{a}^{2}$
$2 \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xy}=\mathrm{a}^{2}$
check $\mathrm{h}^{2}-\mathrm{ab} < 0$
ellipse
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