What is the acute angle between the curves xy=2 and `y^2=4x` at their point of intersection ?

Options

A. `pi/3`
B. `pi/6`
C. `tan^(-1)(3)`
D. `+-tan^(-1)(3)`

Answer

To find the acute angle between the curves \(xy = 2\) and \(y^2 = 4x\) at their point of intersection, we can follow these steps:

### Step 1: Find the point of intersection
We have the two equations:
1. \(xy = 2\)
2. \(y^2 = 4x\)

From the first equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{2}{x} \]

Now, substitute this expression for \(y\) into the second equation:
\[ \left(\frac{2}{x}\right)^2 = 4x \]
\[ \frac{4}{x^2} = 4x \]

Multiplying both sides by \(x^2\) (assuming \(x \neq 0\)):
\[ 4 = 4x^3 \]
\[ x^3 = 1 \quad \Rightarrow \quad x = 1 \]

Now substituting \(x = 1\) back into the first equation to find \(y\):
\[ y = \frac{2}{1} = 2 \]

Thus, the point of intersection is \((1, 2)\).

### Step 2: Find the slopes of the curves at the point of intersection
**For the first curve \(xy = 2\):**
Rearranging gives:
\[ y = \frac{2}{x} \]

Differentiating with respect to \(x\):
\[ \frac{dy}{dx} = -\frac{2}{x^2} \]

At \(x = 1\):
\[ \frac{dy}{dx} = -\frac{2}{1^2} = -2 \]

**For the second curve \(y^2 = 4x\):**
Differentiating implicitly:
\[ 2y \frac{dy}{dx} = 4 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

At the point \((1, 2)\):
\[ \frac{dy}{dx} = \frac{2}{2} = 1 \]

### Step 3: Calculate the acute angle between the curves
Let \(m_1 = -2\) (slope of the first curve) and \(m_2 = 1\) (slope of the second curve). The formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values:
\[ \tan \theta = \left| \frac{-2 - 1}{1 + (-2)(1)} \right| = \left| \frac{-3}{1 - 2} \right| = \left| \frac{-3}{-1} \right| = 3 \]

Thus, we have:
\[ \theta = \tan^{-1}(3) \]

### Conclusion
The acute angle between the curves at their point of intersection is:
\[ \theta = \tan^{-1}(3) \]