Question

Angle between the curves $x^{2}y=1-y$ and $x^3=2(1-y)$ is

• A. ${\tan }^{-1}4$
• B. ${\tan }^{-1}\frac{3}{4}$
• C. ${\tan }^{-1}\frac{4}{7}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C ${\tan }^{-1}\frac{4}{7}$

$x^{2}y=1-y\dots \dots \left(1\right)$
$x^3=2(1-y)\dots \dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we have -
$x^3=2x^{2}y$
$\Rightarrow x=0,x=2y$
From $\left(1\right)$ and $\left(2\right)$
When $x=0,y=1$
When $x=2y\Rightarrow x^2\times \frac{x}{2}=1-\frac{x}{2}$
$\Rightarrow x^3=2-x\Rightarrow x^3+x-2=0$
$\Rightarrow (x-1)(x^2+x+2)=0$
$\therefore x=1\Rightarrow y=\frac{1}{2}$
$\therefore$ Points of intersection are $(0,1)$ and $(1,\frac{1}{2})$
Angle between the curves = Angle between the tangents at the point of intersection
$\begin{array}{cc}{x}^{2}y=1-y& x^3=2(1-y)\\ \text{Differentiating}& \text{Differentiating}\\ x^2{y}^{{}^{\prime }}+y\left(2x\right)=-y^{{}^{\prime }}& 3x^2=-2y^{{}^{\prime }}\\ \Rightarrow y^{{}^{\prime }}=\frac{-2xy}{(1+x^2)}& \Rightarrow y^{{}^{\prime }}=\frac{-3}{2}{x}^2\\ \text{At (0, 1)}m=0& \text{At (0, 1)}m=0\end{array}$
At $(0,1)$ $m_1=0$ $m_2=0\Rightarrow \theta =0$
At $(1,\frac{1}{2})$ $m_1=\frac{-2\times \frac{1}{2}}{2}=\frac{-1}{2}$ $m_2=\frac{-3}{2}$
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$\theta ={\tan }^{-1}\left(\frac{\frac{-1}{2}+\frac{3}{2}}{1+\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)}\right)={\tan }^{-1}\left(\frac{4}{7}\right)$