Question

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

Answer 1

Given curves,

$2x=y^2$ .....$\left(1\right)$

$2xy=k$ .....$\left(2\right)$

From $\left(1\right)$, $x=\frac{y^2}{2}$

Substitute this value of $x$ in $\left(2\right)$, we get

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

$\Rightarrow y^3=k$

$\Rightarrow y=(k{)}^{\frac{1}{3}}$

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

So the two curves intersect at $(\frac{(k{)}^{\frac{2}{3}}}{2},(k{)}^{\frac{1}{3}})$

Now, Consider $2x=y^2$

Differentiate w.r.t $x$

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

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{(k{)}^{\frac{1}{3}}}=(k{)}^{\frac{-1}{3}}$

Let this $\frac{dy}{dx}=m_1$

$\Rightarrow m_1=(k{)}^{\frac{-1}{3}}$

Now, Consider $2xy=k$

Differentiate w.r.t $x$

$\Rightarrow 2y+2x\frac{dy}{dx}=0$

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

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

$\Rightarrow \frac{dy}{dx}=\frac{-(k{)}^{\frac{1}{3}}}{\frac{(k{)}^{\frac{2}{3}}}{2}}$

On simplifying we get,

$\Rightarrow \frac{dy}{dx}=-2(k{)}^{\frac{-1}{3}}$

Let this $\frac{dy}{dx}=m_2$

$\Rightarrow m_2=-2(k{)}^{\frac{-1}{3}}$

Since these 2 curves cut at Right angles,

$\Rightarrow m_1{m}_2=-1$

$\Rightarrow (k{)}^{\frac{-1}{3}}\times (-2k^{\frac{-1}{3}})=-1$

On simplifying we get,

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

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

cubing on both sides, we get

$\Rightarrow k^{-2}=\frac{1}{8}$

$\Rightarrow k^2=8$

Therefore, $\Rightarrow k=2\sqrt{2}$

Hence, the condition that the 2 curves intersect orthogonally is $k=2\sqrt{2}$