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(A) Given that $BOAC$ is a rectangle in the $XY$-plane with $O(0,0)$ as the origin and points $A, B$ lying on the parabola $y=x^2$.Let $A = (t_1, t_1^

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$y=x^2+2$

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(A) Given that $BOAC$ is a rectangle in the $XY$-plane with $O(0,0)$ as the origin and points $A, B$ lying on the parabola $y=x^2$.Let $A = (t_1, t_1^2)$ and $B = (t_2, t_2^2)$.Since $BOAC$ is a rectangle,the diagonals $OA$ and $BC$ bisect each other at the same midpoint,and the sides $OA$ and $OB$ are perpendicular.The slope of $OA$ is $m_1 = \frac{t_1^2 - 0}{t_1 - 0} = t_1$.The slope of $OB$ is $m_2 = \frac{t_2^2 - 0}{t_2 - 0} = t_2$.Since $OA \perp OB$,we have $m_1 \cdot m_2 = -1$,which implies $t_1 t_2 = -1$.Let $C = (h, k)$. Since $BOAC$ is a rectangle,the vector $\vec{OC} = \vec{OA} + \vec{OB}$.Thus,$h = t_1 + t_2$ and $k = t_1^2 + t_2^2$.We can express $k$ in terms of $h$ as follows:$k = (t_1 + t_2)^2 - 2t_1 t_2$$k = h^2 - 2(-1)$$k = h^2 + 2$.Therefore,the locus of $C(h, k)$ is $y = x^2 + 2$.

Suppose $BOAC$ is a rectangle in the $XY$-plane where $O$ is the origin and $A, B$ lie on the parabola $y=x^2$. Then,$C$ must lie on the curve

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