The angle of intersection between the curves `y^2=4x" and "x^2=32y` at point (16,8) is

Options

A. `tan^(-1)(3/5)`
B. `tan^(-1)(4/5)`
C. `pi `
D. `(pi)/(2)`

Answer

To find the angle of intersection between the curves \(y^2 = 4x\) and \(x^2 = 32y\) at the point \((16, 8)\), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ y^2 = 4x \]
Differentiating both sides with respect to \(x\):
\[ 2y \frac{dy}{dx} = 4 \]
Thus, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 2: Find the slope \(m_1\) at the point \((16, 8)\)
Substituting \(y = 8\) into the derivative:
\[ m_1 = \frac{2}{8} = \frac{1}{4} \]

### Step 3: Differentiate the second curve
The second curve is given by:
\[ x^2 = 32y \]
Differentiating both sides with respect to \(x\):
\[ 2x = 32 \frac{dy}{dx} \]
Thus, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2x}{32} = \frac{x}{16} \]

### Step 4: Find the slope \(m_2\) at the point \((16, 8)\)
Substituting \(x = 16\) into the derivative:
\[ m_2 = \frac{16}{16} = 1 \]

### Step 5: Use the formula for the angle between two curves
The angle \(\theta\) between two curves 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 \(m_1 = \frac{1}{4}\) and \(m_2 = 1\):
\[ \tan \theta = \left| \frac{\frac{1}{4} - 1}{1 + \frac{1}{4} \cdot 1} \right| \]
Calculating the numerator:
\[ \frac{1}{4} - 1 = \frac{1 - 4}{4} = \frac{-3}{4} \]
Calculating the denominator:
\[ 1 + \frac{1}{4} = \frac{4 + 1}{4} = \frac{5}{4} \]
Thus:
\[ \tan \theta = \left| \frac{-\frac{3}{4}}{\frac{5}{4}} \right| = \left| -\frac{3}{5} \right| = \frac{3}{5} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \]

### Final Answer
The angle of intersection between the curves at the point \((16, 8)\) is:
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \]