The curve $x = y^2$ and $xy = k$ cut orthogonally, then $k^2$ is equal to:

When two curves "cut orthogonally," they intersect at right angles (90°). For curves to be perpendicular at their meeting point, the product of their slopes must equal -1.

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For the curve $x = y^2$, differentiate both sides with respect to $x$:

$1 = 2y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{2y}$

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For the curve $xy = k$, differentiate both sides with respect to $x$:

$y + x\frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{y}{x}$

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For orthogonal curves, the product of slopes equals -1:

$\frac{1}{2y} \times \left(-\frac{y}{x}\right) = -1$

$-\frac{y}{2yx} = -1$

$-\frac{1}{2x} = -1$

$\frac{1}{2x} = 1$

$x = \frac{1}{2}$

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The point of intersection lies on $x = y^2$:

$\frac{1}{2} = y^2$

$y = \pm\frac{1}{\sqrt{2}}$

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From the equation $xy = k$:

$k = \frac{1}{2} \times \frac{1}{\sqrt{2}}$

$k = \frac{1}{2\sqrt{2}}$

Therefore:

$k^2 = \left(\frac{1}{2\sqrt{2}}\right)^2$

$k^2 = \frac{1}{4 \times 2}$

$k^2 = \frac{1}{8}$