The angle of intersection of the curve `y = x^2` and `6y=7-x^2` at (1,1) is

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. None of these

Answer

To find the angle of intersection of the curves \( y = x^2 \) and \( 6y = 7 - x^2 \) at the point \( (1, 1) \), we will follow these steps:

### Step 1: Find the slopes of the tangents to the curves at the point of intersection.

1. **For the first curve** \( y = x^2 \):
- Differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2x \]
- Evaluate at the point \( (1, 1) \):
\[ m_1 = 2(1) = 2 \]

2. **For the second curve** \( 6y = 7 - x^2 \):
- First, rewrite it as \( y = \frac{7 - x^2}{6} \).
- Differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{-2x}{6} = -\frac{x}{3} \]
- Evaluate at the point \( (1, 1) \):
\[ m_2 = -\frac{1}{3} \]

### Step 2: Use the formula for the angle of intersection.

The formula for the tangent of the angle \( \theta \) between two curves with slopes \( m_1 \) and \( m_2 \) is:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

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

### Step 3: Simplify the expression.

1. Calculate the numerator:
\[ 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \]

2. Calculate the denominator:
\[ 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \]

3. Now, substitute back into the formula:
\[ \tan \theta = \frac{\frac{7}{3}}{\frac{1}{3}} = 7 \]

### Step 4: Find the angle \( \theta \).

To find \( \theta \):
\[ \theta = \tan^{-1}(7) \]

### Final Result:
The angle of intersection of the curves \( y = x^2 \) and \( 6y = 7 - x^2 \) at the point \( (1, 1) \) is \( \tan^{-1}(7) \).

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