For what value(s) of k do the tangents of two curves x = y² and xy = k cut at right angles?

For what value(s) of $k$ do the tangents of two curves $x = y^2$ and $xy = k$ cut at right angles?

a$k = \pm \frac{1}{4}$
b$k = \pm 1$
c$k = 0$
d$k = \pm \frac{1}{2\sqrt{2}}$

Answer

Official answer key

Correct option: d

Answer: (d) $k = \pm\frac{1}{2\sqrt{2}}$ For $x = y^2$: $\frac{dy}{dx} = \frac{1}{2y}$. For $xy = k$: $\frac{dy}{dx} = -\frac{y}{x}$. Perpendicular tangents: $\frac{1}{2y}\cdot\left(-\frac{y}{x}\right) = -1$, so $x = \frac{1}{2}$ and $y^2 = \frac{1}{2}$. Then $k^2 = x^2y^2 = \frac{1}{8}$, so $k = \pm\frac{1}{2\sqrt{2}}$.