Question

If the curves $x=y^4$ and $xy=k$ cut at right angles, then $(4k{)}^6$ is equal to

Answer 1

$4y^3\frac{dy}{dx}=1\&x\frac{dy}{dx}+y=0$
At $(x_1,y_1)$
$m_1=\frac{1}{4y_1^3},m_2=\frac{-y_1}{x_1}$
For curves to intersect at right angles $m_1{m}_2=-1$
$\Rightarrow \frac{1}{4.y_1^3}\times \frac{-y_1}{x_1}=-1$
$\Rightarrow \frac{1}{4.y_1^6}=1(\because x_1=y_1^4)$
$y_1^6=\frac{1}{4}$
Now, $x_1{y}_1=k\Rightarrow k=y_1^5$
$\Rightarrow k^6=y_1^{30}={\left(\frac{1}{4}\right)}^5$
$\therefore (4k{)}^6=4^6\times k^6=4$