Rectangle Meets Parabola

Solution 1:

I've discovered a much better solution than the one I originally gave. If the coordinates for $A$ are $(x_A, x_{A}^{2})$ , then the coordinates for $C$ must be $(-\dfrac{1}{x_A}, \frac{1}{x_{A}^{2}})$ . The vectors to these points from the origin must be perpendicular since this is a rectangle. Therefore, their dot product must be 0. Indeed, $\langle x_A, x_{A}^{2} \rangle \cdot \langle-\dfrac{1}{x_A}, \frac{1}{x_{A}^{2}} \rangle = 0$ . By the parallelogram rule for vectors, vector $OA +$ vector $OC =$ vector $OB$ . Therefore, $B = \langle x_A, x_{A}^{2} \rangle + \langle -\dfrac{1}{x_A}, \frac{1}{x_{A}^{2}} \rangle$ , thereby making the coordinates of $B(x_A - \dfrac{1}{x_A}, x_{A}^{2} + \dfrac{1}{x_{A}^{2}}).$

Since $x_{B}^{2} = \left(x_A - \dfrac{1}{x_A}\right)^2 = x_{A}^{2} - 2 + \dfrac{1}{x_{A}^2} = y_B - 2$ , we have that $\boxed{y_B = x_{B}^{2} + 2}.$

Solution 2:

Denote points $A(x_A, y_A)$ and $C(x_C, y_C)$ . Moreover, because these points sit on $y = x^2$ , we know that $y_A = x_{A}^{2}$ and $y_C = x_{C}^{2}$ .

The side of the rectangle OA is contained in the line given by $y = x_Ax$ , while the side of the rectangle OC is contained in the line given by $y = -\dfrac{1}{x_A}x$ because the sides of a rectangle are perpendicular.
(Also, since $(x_C, x_{C}^{2})$ sits on the line $y = -\dfrac{1}{x_A}x$ , we have that $x_{C}^{2} = -\dfrac{x_C}{x_A} \longrightarrow \color{#D61F06}{x_Ax_C = -1}$ )

Now, we construct the lines containing sides AB and BC using the slopes from the first two lines, but now passing through A and C (in the appropriate equations for perpendicularity). That is, the line containing AB will be given by

$y - y_A = -\frac{1}{x_A}(x - x_A) \longrightarrow y = -\frac{x}{x_A} + 1 + y_A = -\frac{x}{x_A} + 1 + x_{A}^{2}$

and the line containing BC will be given by

$y - y_C = x_A(x - x_C) \longrightarrow y = (x_A)x - x_Ax_C + y_C = (x_A)x - x_Ax_C + x_{C}^{2}$

In order to determine point B, we set the $y$ -values of these lines equal to each other, thus giving

$(x_A)x - x_Ax_C + x_{C}^{2} = -\frac{x}{x_A} + 1 + x_{A}^{2}$

$(x_A)x - (\color{#D61F06}{-1}) + x_{C}^{2} = -\frac{x}{x_A} + 1 + x_{A}^{2}$

From here, a bit of algebra will show that the $x$ -coordinate of B (that is, the intersection of the two lines that contain AB and BC) is $x_B = x_A - \dfrac{1}{x_A}$ and the corresponding $y$ -value from this intersection is $y_B = x_{A}^{2} + \dfrac{1}{x_{A}^{2}}$ .

Since $x_{B}^{2} = \left(x_A - \dfrac{1}{x_A}\right)^2 = x_{A}^{2} - 2 + \dfrac{1}{x_{A}^2} = y_B - 2$ , we have that $\boxed{y_B = x_{B}^{2} + 2}.$