Prove that the curves

The equations of the given curves are given as $x=y^{2}$ and $x y=k$.

Putting $x=y^{2}$ in $x y=k$, we get:

$y^{3}=k \Rightarrow y=k^{\frac{1}{3}}$

$\therefore x=k^{\frac{2}{3}}$

Thus, the point of intersection of the given curves is $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$.

Differentiating $x=y^{2}$ with respect to $x$, we have:

$1=2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$

Therefore, the slope of the tangent to the curve $x=y^{2}$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is $\left.\frac{d y}{d x}\right]\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)=\frac{1}{2 k^{\frac{1}{3}}}$.

On differentiating $x y=k$ with respect to $x$, we have:

$x \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

$\therefore$ Slope of the tangent to the curve $x y=k$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is given by,

$\left.\left.\frac{d y}{d x}\right]\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)=\frac{-y}{x}\right]\left(k^{\frac{2}{3}, k^{\frac{1}{3}}}\right)=-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}}=\frac{-1}{k^{\frac{1}{3}}}$

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ are

perpendicular to each other.

This implies that we should have the product of the tangents as − 1.

Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is $-1 .$

i.e. $\left(\frac{1}{2 k^{\frac{1}{3}}}\right)\left(\frac{-1}{k^{\frac{1}{3}}}\right)=-1$

$\Rightarrow 2 k^{\frac{2}{3}}=1$

$\Rightarrow\left(2 k^{\frac{2}{3}}\right)^{3}=(1)^{3}$

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

Hence, the given two curves cut at right angels if $8 k^{2}=1$.