Question

Find the angle of intersection of the following curve  x² = 27y and y² = 8x ?

Answer 1

$$\text{ Given curves are },$$

$$x^2=27y...\left(1\right)$$

$$y^2=8x...\left(2\right)$$

$$\text{ From }(2)\text{ we get }$$

$$x=\frac{y^2}{8}$$

$$\text{ Substituting this in }(1),$$

$${\left(\frac{y^2}{8}\right)}^2=27y$$

$$\Rightarrow y^4=1728y$$

$$\Rightarrow y(y^3-{12}^3)=0$$

$$\Rightarrow y=0ory=12$$

$$\text{ Substituting the values of y in (2), we get },$$

$$\Rightarrow x=0orx=18$$

$$\Rightarrow (x,y)=(0,0),(18,12)$$

$$\text{ Differentiating (1) w.r.t.x },$$

$$2x=27\frac{dy}{dx}$$

$$\Rightarrow \frac{dy}{dx}=\frac{2x}{27}...\left(3\right)$$

$$\text{ Differentiating (2) w.r.t.x },$$

$$2y\frac{dy}{dx}=8$$

$$\Rightarrow \frac{dy}{dx}=\frac{4}{y}...\left(4\right)$$

$$\text{ Case }-1:(x,y)=(0,0)$$

$$\text{ From }\left(4\right)\text{ we have,}{m}_2\text{ is undefined }$$

$$\therefore \text{ We cannot find }\theta$$

$$\text{ Case -}2:(x,y)=(18,12)$$

$$\text{ From }\left(3\right)\text{ we have },m_1=\frac{36}{27}=\frac{4}{3}$$

$$\text{ From }\left(4\right)\text{ we have },m_2=\frac{4}{12}=\frac{1}{3}$$

$$\text{ Now },$$

$$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}\right|=\frac{9}{13}$$

$$\Rightarrow \theta ={\tan }^{-1}\left(\frac{9}{13}\right)$$