Question

The angle between the curves $y=2x$ and $x^3+y^3=3xy$ at the point other than the origin is

Answer 1

Step 1- Finding the point

Given two curves are:

$y=2xeq\left(1\right)$

$x^3+y^3=3xyeq\left(2\right)$

Putting the value of $eq\left(1\right)$ in $eq\left(2\right)$
$$\begin{gathered} \Rightarrow x^3+{\left(2x\right)}^3=3\times x\times 2x \\ \Rightarrow x^3+8x^3=6x^2 \\ \Rightarrow 9x^3=6x^2 \\ \Rightarrow 9x^3-6x^2=0 \\ \Rightarrow 3x^2(3x-2)=0 \\ \Rightarrow x=0orx=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \because y=2x \\ whenx=0\Rightarrow y=0 \\ whenx=\frac{2}{3}\Rightarrow y=\frac{4}{3} \end{gathered}$$

So, the points are $\left(0,0\right)or\left(\frac{2}{3},\frac{4}{3}\right)$

Since the point is other than the origin

$\Rightarrow (x,y)=\left(\frac{2}{3},\frac{4}{3}\right)$

Step 2 - finding the slope

differentiate the equations

$$\begin{gathered} \because y=2x \\ \Rightarrow \frac{dy}{dx}=2 \\ \Rightarrow m_1=2 \end{gathered}$$

And,

$$\begin{gathered} \because x^3+\hspace{0.17em}{y}^3=3xy \\ \Rightarrow 3x^2+3y^2\frac{dy}{dx}=3y+3x\frac{dy}{dx} \\ \Rightarrow \left(3y^2-3x\right)\frac{dy}{dx}=3y-3x^2 \\ \Rightarrow \frac{dy}{dx}=\frac{3(y-x^2)}{3(y^2-x)} \\ \Rightarrow \frac{dy}{dx}=\frac{y-x^2}{y^2-x} \end{gathered}$$

Now, substitute $x=\frac{2}{3}$ and $y=\frac{4}{3}$we get,

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{{\displaystyle \frac{4}{3}}-{\displaystyle \frac{4}{9}}}{{\displaystyle \frac{16}{9}}-{\displaystyle \frac{2}{3}}} \\ \Rightarrow \frac{dy}{dx}=\frac{4}{5} \end{gathered}$$

$\Rightarrow m_2=\frac{4}{5}$

Step 3 - Angle between the curve

If $m_1>m_2$

$$\begin{gathered} \because \tan \left(\theta \right)=\frac{m_1-m_2}{1+m_1{m}_2} \\ \Rightarrow \tan \left(\theta \right)=\frac{2-{\displaystyle \frac{4}{5}}}{1+2\times {\displaystyle \frac{4}{3}}} \\ \Rightarrow \tan \left(\theta \right)=\frac{{\displaystyle \frac{6}{5}}}{{\displaystyle \frac{13}{5}}} \\ \Rightarrow \tan \left(\theta \right)=\frac{6}{13} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{6}{13}\right) \end{gathered}$$

Hence, the angle between the curve is ${\tan }^{-1}\left(\frac{6}{13}\right)$