Step 1: Understanding the Concept:
The angle between two intersecting curves is defined as the angle between their tangent lines at the point of intersection. We first need to algebraically find the point(s) where they cross. Then, we find the derivatives (slopes of tangents) of both curves at that specific point. Finally, we use the angle formula for two lines.
Step 2: Key Formula or Approach:
1. Solve the system of equations $xy = 6$ and $x^2y = 12$ to find the intersection point $(x_0, y_0)$.
2. Differentiate each curve equation to find general slope expressions $y_1'$ and $y_2'$.
3. Evaluate slopes at intersection: $m_1 = y_1'(x_0)$ and $m_2 = y_2'(x_0)$.
4. Use the angle formula: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Step 3: Detailed Explanation:
First, find the point of intersection.
Curve 1: $xy = 6 \implies y = \frac{6}{x}$
Curve 2: $x^2y = 12$
Substitute $y$ from Curve 1 into Curve 2:
\[ x^2 \left(\frac{6}{x}\right) = 12 \]
Assuming $x \neq 0$ (from the domain of hyperbola), simplify:
\[ 6x = 12 \implies x = 2 \]
Substitute $x = 2$ back to find $y$:
\[ y = \frac{6}{2} = 3 \]
The intersection point is $(2, 3)$.
Next, find the slopes of the tangents at $(2, 3)$.
For Curve 1 ($xy = 6$), differentiate implicitly or explicitly. Let's do explicitly: $y = 6x^{-1}$
\[ \frac{dy}{dx} = -6x^{-2} = \frac{-6}{x^2} \]
Slope $m_1$ at $x = 2$:
\[ m_1 = \frac{-6}{2^2} = \frac{-6}{4} = -\frac{3}{2} \]
For Curve 2 ($x^2y = 12$), explicit form: $y = 12x^{-2}$
\[ \frac{dy}{dx} = -24x^{-3} = \frac{-24}{x^3} \]
Slope $m_2$ at $x = 2$:
\[ m_2 = \frac{-24}{2^3} = \frac{-24}{8} = -3 \]
Now, use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substitute $m_1 = -3/2$ and $m_2 = -3$:
\[ \tan \theta = \left| \frac{-\frac{3}{2} - (-3)}{1 + \left(-\frac{3}{2}\right)(-3)} \right| \]
\[ \tan \theta = \left| \frac{-\frac{3}{2} + 3}{1 + \frac{9}{2}} \right| \]
Find common denominators:
\[ \tan \theta = \left| \frac{\frac{-3 + 6}{2}}{\frac{2 + 9}{2}} \right| = \left| \frac{\frac{3}{2}}{\frac{11}{2}} \right| \]
\[ \tan \theta = \left| \frac{3}{2} \times \frac{2}{11} \right| = \frac{3}{11} \]
Therefore, the angle is $\theta = \tan^{-1}\left(\frac{3}{11}\right)$.
Step 4: Final Answer:
The angle between the curves is $\tan^{-1} \frac{3}{11}$.