Consider a rigid square ABCD as in the figure | vedclass

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(D) Let the side length of the square $ABCD$ be $a$. Let the coordinates of $A$ be $(x_A, 0)$ and $B$ be $(0, y_B)$. Let the angle $\angle OAB = 	het

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an ellipse which is not a circle

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(D) Let the side length of the square $ABCD$ be $a$. Let the coordinates of $A$ be $(x_A, 0)$ and $B$ be $(0, y_B)$. Let the angle $\angle OAB = 	heta$. Then $A = (a \cos 	heta, 0)$ and $B = (0, a \sin 	heta)$.Since $ABCD$ is a square,the vector $\vec{BC}$ is perpendicular to $\vec{AB}$ and has the same length $a$. The vector $\vec{AB} = (-a \cos 	heta, a \sin 	heta)$. Rotating this by $90^\circ$ gives $\vec{BC} = (a \sin 	heta, a \cos 	heta)$.Thus,$C = B + \vec{BC} = (0 + a \sin 	heta, a \sin 	heta + a \cos 	heta) = (a \sin 	heta, a(\sin 	heta + \cos 	heta))$.Let $C = (x, y)$. Then $x = a \sin 	heta$ and $y = a \sin 	heta + a \cos 	heta$.From the first equation,$\sin 	heta = \frac{x}{a}$. Then $\cos 	heta = \sqrt{1 - (\frac{x}{a})^2} = \frac{\sqrt{a^2 - x^2}}{a}$.Substituting into the second equation: $y = x + \sqrt{a^2 - x^2}$.$y - x = \sqrt{a^2 - x^2}$.Squaring both sides: $(y - x)^2 = a^2 - x^2$.$y^2 - 2xy + x^2 = a^2 - x^2$.$2x^2 + y^2 - 2xy = a^2$.This is the equation of an ellipse,as the discriminant $B^2 - 4AC = (-2)^2 - 4(2)(1) = 4 - 8 = -4 < 0$. Since the coefficients of $x^2$ and $y^2$ are not equal,it is not a circle.

Consider a rigid square $ABCD$ as in the figure with $A$ and $B$ on the $X$ and $Y$-axes,respectively. When $A$ and $B$ slide along their respective axes,the locus of $C$ forms a part of

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