Question

If the curves $x=y^2\mathrm{and}\mathrm{xy}=k$ cut at right angles, then the value of $8k^2$ is .

Answer 1

The given curves are $x=y^2.....\left[1\right]\mathrm{xy}=k......\left[2\right]$
Solving [1] nad [2] we get the point of intersection of the two given curves.
$y^3=k\Rightarrow y=k^{\frac{1}{3}}\therefore x=k^{\frac{2}{3}}$
So the point of intersection of the two curves is $P\left(k^{\frac{2}{3}},k^{\frac{1}{3}}\right)$.
Differentiating [1] both sides w.r.t. x, we get
$1=2y\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2y}\Rightarrow m_1={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_P=\frac{1}{2k^{\frac{1}{3}}}$
Differentiating [2] both sides w.r.t. x, we get
$1\cdot y+x\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{y}{x}\Rightarrow m_2={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_P=-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}}\Rightarrow m_2=-\frac{1}{k^{\frac{1}{3}}}.$For the curves [1] and [2] to cut at right angles at P, we must have
$m_1\times m_2=-1\Rightarrow \frac{1}{2k^{\frac{1}{3}}}\times -\frac{1}{k^{\frac{1}{3}}}=-1\Rightarrow 2k^{\frac{2}{3}}=1\Rightarrow {\left(2k^{\frac{2}{3}}\right)}^3=1^3\Rightarrow 8k^2=1.$