Angle of intersection of the curve `x^2 =32y` and `y^2 =4x` at.the point (16, 8) is

Options

A. `60^@`
B. `90^@`
C. `tan^(-1)(4//3)`
D. `tan^(-1)(3//5)`

Answer

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

### Step 1: Find the derivatives of the curves
We need to find the slopes of the tangents to both curves at the point \( (16, 8) \).

**For the first curve \( C_1: x^2 = 32y \)**:
1. Differentiate both sides with respect to \( x \):
\[ 2x = 32 \frac{dy}{dx} \]
2. Rearranging gives:
\[ \frac{dy}{dx} = \frac{2x}{32} = \frac{x}{16} \]
3. Substitute \( x = 16 \):
\[ m_1 = \frac{16}{16} = 1 \]

**For the second curve \( C_2: y^2 = 4x \)**:
1. Differentiate both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 4 \]
2. Rearranging gives:
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]
3. Substitute \( y = 8 \):
\[ m_2 = \frac{2}{8} = \frac{1}{4} \]

### Step 2: Use the angle formula
The angle \( \theta \) between the two curves can be calculated using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_1 = 1 \) and \( m_2 = \frac{1}{4} \):
\[ \tan \theta = \left| \frac{1 - \frac{1}{4}}{1 + 1 \cdot \frac{1}{4}} \right| \]
Calculating the numerator:
\[ 1 - \frac{1}{4} = \frac{3}{4} \]
Calculating the denominator:
\[ 1 + \frac{1}{4} = \frac{5}{4} \]
Thus,
\[ \tan \theta = \left| \frac{\frac{3}{4}}{\frac{5}{4}} \right| = \frac{3}{5} \]

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

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