Question

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

Answer 1

Given curves are, $x=y^2\cdots \left(i\right)$
$xy=k\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
we get,
$\Rightarrow y^3=k$
$\Rightarrow y=k^{\frac{1}{3}}$
From $\left(i\right)$,we get
$x=k^{\frac{2}{3}}$
Thus, point of intersection of curve is $(k^{\frac{2}{3}},k^{\frac{1}{3}})$
Slope of tangent for $x=y^2$
Differentiating w.r.t. $x,\frac{dx}{dx}=\frac{d\left(y^2\right)}{dx}$
$\Rightarrow 1=2y\times \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$
Slope of tangent at $(k^{\frac{2}{3}},k^{\frac{1}{3}})$ is $\frac{dy}{dx}=\frac{1}{2\left(k^{\frac{1}{3}}\right)}=\frac{1}{2k^{\frac{1}{3}}}\cdots \left(iii\right)$
Slope of tangent for $xy=k$
Differentiating w.r.t. $x,\frac{d\left(xy\right)}{dx}=0$
$\Rightarrow y+x\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-y}{x}$
Slope of tangent at $(k^{\frac{2}{3}},k^{\frac{1}{3}})$ is $\frac{dy}{dx}=\frac{-k^{\frac{1}{3}}}{k^{\frac{2}{3}}}=\frac{-1}{k^{\frac{1}{3}}}\cdots \left(iv\right)$
For the curve to cut at right angles, the tangents must be perpendicular at point of intersection.
Product of the slopes of the tangents $=-1$
$\frac{1}{2k^{\frac{1}{3}}}\times \frac{-1}{k^{\frac{1}{3}}}=-1$
$\Rightarrow \frac{1}{2k^{\frac{2}{3}}}=1$
$\Rightarrow 2k^{\frac{2}{3}}=1$
$\Rightarrow k^{\frac{2}{3}}=\frac{1}{2}$
Taking cube on both sides,
$\Rightarrow {\left(k^{\frac{2}{3}}\right)}^3={\left(\frac{1}{2}\right)}^3$
$\Rightarrow k^2=\frac{1}{8}$
$\Rightarrow 8k^2=1$