Prove that the curves $x = {y^2}and\,\,xy = k$ cut at right angles, if $8{k^2} = 1.$

We have,
$x = {y^2}$ …(i)
and $xy = k$ …(ii)

Solving (i) \&(ii), we get ${y^3} = k \Rightarrow y = {k^{1/3}}$
Substituting this value of y in (i), we get $x = {({k^{1/3}})^2} = {k^{2/3}}$

Therefore, (i) and (ii) intersect at the point $({k^{2/3}},\;{k^{1/3}})$ .

Differentiating (i) w.r.t. $x$, we get
$1 = 2y\cfrac{{dy}}{{dx}} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{2y}}$

Slope of tangent to (i) at $({k^{2/3}},\;{k^{1/3}}) = \cfrac{1}{{2{k^{1/3}}}}$ …(iii)
From (ii), $y = \cfrac{k}{x}$

Differentiating (ii) wr.t.$x$, we get $\cfrac{{dy}}{{dx}} = - \cfrac{k}{{{x^2}}}$

Therefore the slope of tangent to (ii) at $({k^{2/3}},\;{k^{1/3}})$
$= - \cfrac{k}{{{{({k^{2/3}})}^2}}} = - \cfrac{1}{{{k^{1/3}}}}$ …(iv)

The two curves cut at right angles (i.e., orthogonally) at $({k^{2/3}},\;{k^{1/3}})$ , if product of slopes of their tangents $= - 1$

$\Rightarrow \left( {\cfrac{1}{{2{k^{1/3}}}}} \right)\left( { - \cfrac{1}{{{k^{1/3}}}}} \right) = - 1 \Rightarrow 1 = 2{k^{2/3}} \Rightarrow 1 = 8{k^2}.$