For what value(s) of do the tangents of two curves and cut at right angles?

(d) $k = \pm \frac{1}{2\sqrt{2}}$ For $x = y^2$, differentiating gives $1 = 2y\frac{dy}{dx} \Rightarrow m_1 = \frac{1}{2y}$. For $xy = k$, differentiating gives $x\frac{dy}{dx} + y = 0 \Rightarrow m_2 = -\frac{y}{x}$. Since the tangents are perpendicular, $m_1 m_2 = -1 \Rightarrow \left(\frac{1}{2y}\right)\left(-\frac{y}{x}\right) = -1 \Rightarrow -\frac{1}{2x} = -1 \Rightarrow x = \frac{1}{2}$. Since $x = y^2$, we have $y^2 = \frac{1}{2}$. Then $k^2 = x^2 y^2 = \left(\frac{1}{4}\right)\left(\frac{1}{2}\right) = \frac{1}{8} \Rightarrow k = \pm \frac{1}{2\sqrt{2}}$.