#### Find the condition that the curves $2x = y^2$ and 2xy = k intersect orthogonally.

Given: two curves   $2x = y^2$ and 2xy = k

To find: to track the condition where both the curves intersect orthogonally

Explanation: Given 2xy = k

$\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{2 \mathrm{x}} \ldots \ldots (i)$
Put in the value of y in another curve equation, i.e., $2 x=y^{2},$ we get

$2 \mathrm{x}=\left(\frac{\mathrm{k}}{2 \mathrm{x}}\right)^{2}$
$\\\Rightarrow 2 \mathrm{x}=\frac{\mathrm{k}^{2}}{4 \mathrm{x}^{2}} \\\Rightarrow \mathrm{x}^{3}=\frac{\mathrm{k}^{2}}{8}$
Putting both the sides under cube root, we get

$\Rightarrow \mathrm{x}=\frac{\mathrm{k}^{\frac{2}{3}}}{2} \ldots \ldots (ii)$

Substituting equation (ii) in equation (i), we get

$\Rightarrow y=\frac{k}{2\left ( \frac{k^{\frac{2}{3}}}{2} \right )}$

\begin{aligned} &\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mathrm{k}^{\frac{2}{3}}}\\ &\Rightarrow \mathrm{y}=\mathrm{k}^{1-\frac{2}{3}}\\ &\Rightarrow \mathrm{y}=\mathrm{k}^{\frac{1}{3}}\\ & \end{aligned}

$\text { Thus, } \left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$ is the point of intersection of two curves

Now given $2x = y^2$

After differentiating the equation for x, we get

$\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}} \Rightarrow 2=2 \mathrm{y} \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{y}}$

After tracking the value of differentiation at the point of intersection, i.e., at$\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$  we
get

$\left ( \frac{dy}{dx} \right )$$\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$$=\frac{1}{k^{\frac{1}{3}}}=m_1...(ii)$

Also given 2xy = k

Differentiating this with respect to x, we get

\begin{aligned} &\frac{\mathrm{d}(2 \mathrm{xy})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{k})}{\mathrm{dx}}\\ &\Rightarrow 2 \frac{\mathrm{d}(\mathrm{xy})}{\mathrm{dx}}=0\\ &\text { When applying the product rule of differentiation, it results into }\\ &\Rightarrow 2\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)=0\\ &\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0\\ &\Rightarrow x \frac{d y}{d x}=-y\\ &\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \end{aligned}

Then, again, finding the given differentiation value at the point of intersection, i.e., at $\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$, we get

$\left ( \frac{dy}{dx} \right )$$\left(\frac{\mathrm{k}^{\frac{2}{3}}}{2}, \mathrm{k}^{\frac{1}{3}}\right)\\$

$\\=-\frac{k^{\frac{1}{3}}}{\frac{k^{\frac{2}{3}}}{2}} \\ =-2 k^{\frac{1}{3}-\frac{2}{3}} \\ =-2 k^{-\frac{1}{3}}=m_{2} \ldots \text { (iv) }$

However, the orthogonal interaction of two curves occurs if

m1.m2 = -1

Then, Substituting the values from equation (iii) and (iv), we get

\begin{aligned} &\frac{1}{k^{\frac{1}{3}}} \cdot\left(-2 k^{-\frac{1}{3}}\right)=-1\\ &\Rightarrow \mathrm{k}^{-\frac{1}{3}} \cdot\left(-2 \mathrm{k}^{-\frac{1}{3}}\right)=-1\\ &\Rightarrow-2 k^{-\frac{1}{3}-\frac{1}{3}}=-1\\ &\Rightarrow 2 k^{-\frac{2}{3}}=1\\ &\Rightarrow \frac{2}{k^{\frac{2}{3}}}=1\\ &\Rightarrow \mathrm{k}^ \frac{2}{3}=2\\ &\text { Now taking cube on both sides, we get }\\ &\Rightarrow \mathrm{k}^{2}=2^{3}=8\\ &\Rightarrow \mathrm{k}=\pm2 \sqrt 2 \end{aligned}

This condition proves to fulfil the orthogonal interaction point for the two curves.