Question

The angle between the curves $xy = 6$ and $x^2y = 12$ is

• A. $\tan^{-1} \frac{3}{11}$
• B. $\tan^{-1} \frac{11}{3}$
• C. $\tan^{-1} \frac{2}{11}$
• D. $\tan^{-1} \frac{1}{11}$

Hint:

Angle between curves uses slope formula.

Answer 1

The Correct Option is A

Step 1: Find point of intersection. \[ xy=6,\quad x^2y=12 \Rightarrow x=2,\; y=3 \]
Step 2: Find slopes. From $xy=6$: \[ y + x\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = -\frac{3}{2} \] From $x^2y=12$: \[ 2xy + x^2\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = -\frac{2y}{x} = -3 \]
Step 3: Angle between curves. \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{-\frac{3}{2} + 3}{1 + \frac{9}{2}}\right| \] \[ = \frac{\frac{3}{2}}{\frac{11}{2}} = \frac{3}{11} \]
Step 4: Conclusion. \[ {\tan^{-1}\frac{3}{11}} \]