Question

Prove that the curve x = $2y^2$ and xy= K cut at right angle, if $8k^2$ is equals to $1$

Answer 1

$x=y^2$...(1)

$xy=k$..(2)

From (1)

$\left(y^2\right)y=k\Rightarrow y=\sqrt[3]{3k}$

when, $y=\sqrt[3]{3k}\to x=k^{\frac{2}{3}}$

Point of intersection of given curves is $(k^{\frac{2}{3}},\sqrt[3]{3k})$

$x=y^2$

$1=2y\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$

slope of tangent $\frac{dy}{dx}=\frac{1}{2\sqrt[3]{3k}}$...A

$xy=k$

$x\frac{dy}{dx}+y=0$

$\frac{dy}{dx}=\frac{-y}{x}$

slope of tangent $\frac{dy}{dx}=\frac{-\sqrt[3]{3k}}{k^{\frac{2}{3}}}$...B

given curves cut at right angle if and only if tangents are perpendicular to each other.

therefore $A\times B=-1$

$\frac{1}{2\sqrt[3]{3k}}\times \frac{-\sqrt[3]{3k}}{k^{\frac{2}{3}}}=-1$

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

$\Rightarrow 8k^2=1$