find the angle of intersection of the curve xy=6 and `x^2y=12`

Explanation

To find the angle of intersection of the curves given by the equations \(xy = 6\) and \(x^2y = 12\), we will follow these steps:

### Step 1: Find the points of intersection
We start with the two equations:
1. \(xy = 6\) (Equation 1)
2. \(x^2y = 12\) (Equation 2)

From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = \frac{6}{x} \]

Now, substitute this expression for \(y\) into Equation 2:
\[ x^2 \left(\frac{6}{x}\right) = 12 \]
This simplifies to:
\[ 6x = 12 \quad \Rightarrow \quad x = 2 \]

Now, substitute \(x = 2\) back into Equation 1 to find \(y\):
\[ y = \frac{6}{2} = 3 \]

Thus, the curves intersect at the point \((2, 3)\).

### Step 2: Find the slopes of the tangents at the point of intersection
Next, we need to find the derivatives of both equations to determine the slopes of the tangents at the intersection point.

**For Equation 1:**
Differentiating \(xy = 6\) implicitly with respect to \(x\):
\[ x \frac{dy}{dx} + y = 0 \]
Solving for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{y}{x} \]
At the point \((2, 3)\):
\[ \frac{dy}{dx} = -\frac{3}{2} \quad (m_1) \]

**For Equation 2:**
Differentiating \(x^2y = 12\) implicitly with respect to \(x\):
\[ 2xy + x^2 \frac{dy}{dx} = 0 \]
Solving for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{2y}{x} \]
At the point \((2, 3)\):
\[ \frac{dy}{dx} = -\frac{2 \cdot 3}{2} = -3 \quad (m_2) \]

### Step 3: Use the slopes to find the angle of intersection
Let \(\theta\) be the angle of intersection of the two curves. The formula for the tangent of the angle between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

Substituting the values of \(m_1\) and \(m_2\):
\[ \tan \theta = \frac{-\frac{3}{2} - (-3)}{1 + \left(-\frac{3}{2}\right)(-3)} \]
This simplifies to:
\[ \tan \theta = \frac{-\frac{3}{2} + 3}{1 + \frac{9}{2}} = \frac{\frac{3}{2}}{\frac{11}{2}} = \frac{3}{11} \]

### Step 4: Calculate \(\theta\)
Finally, to find \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{3}{11}\right) \]

Thus, the angle of intersection of the curves \(xy = 6\) and \(x^2y = 12\) is:
\[ \theta = \tan^{-1}\left(\frac{3}{11}\right) \]