Question

If $\theta$ is the acute angle between $y^2=x^3$ and $y=2x^2-1$ at $(1,1),$ then $\tan \theta$ is equal to

• A. $\frac{5}{14}$
• B. $\frac{5}{12}$
• C. $\frac{25}{12}$
• D. $\frac{12}{5}$

Answer 1

The correct option is A $\frac{5}{14}$

Let the slope of the tangent to the curves at the point of intersection be $m_1,m_2$
For $y^2=x^3$
Differentiating w.r.t $x$, Slope of tangent is
$\Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{(1,1)}=\frac{3}{2};$
For the curve $y=2x^2-1$
$m_2={\left(\frac{dy}{dx}\right)}_{(1,1)}=4$
Angle between two lines is given by formula
$\theta ={\tan }^{-1}\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$
Putting the value of $m_1,m_2$
$\tan \theta =\left|\frac{4-\frac{3}{2}}{1+4\cdot \frac{3}{2}}\right|=\frac{5}{14}$
$\Rightarrow \tan \theta =\frac{5}{14}$
as $\theta$ is acute, so negative value is rejected