Angle of intersection of the following curves
`y=x^2,6y=7-x^2` is `tan^(-1)7`.

Answer

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

### Step 1: Find the derivatives of both curves.

1. **For the first curve \( y = x^2 \)**:
\[ \frac{dy}{dx} = 2x \]
So, at the point of intersection, the slope \( m_1 = 2x \).

2. **For the second curve \( 6y = 7 - x^2 \)**:
Rearranging gives \( y = \frac{7 - x^2}{6} \).
Now, differentiate:
\[ \frac{dy}{dx} = \frac{-2x}{6} = -\frac{x}{3} \]
So, at the point of intersection, the slope \( m_2 = -\frac{x}{3} \).

### Step 2: Find the points of intersection.

To find the points where the curves intersect, we set \( y = x^2 \) into the second equation:
\[ 6(x^2) = 7 - x^2 \]
This simplifies to:
\[ 7x^2 = 7 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1 \]
Substituting \( x = 1 \) and \( x = -1 \) back into \( y = x^2 \):
- For \( x = 1 \), \( y = 1^2 = 1 \) → point \( (1, 1) \)
- For \( x = -1 \), \( y = (-1)^2 = 1 \) → point \( (-1, 1) \)

Thus, the points of intersection are \( (1, 1) \) and \( (-1, 1) \).

### Step 3: Calculate the angle of intersection.

The formula for the angle \( \theta \) between two curves is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the slopes \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{2x - \left(-\frac{x}{3}\right)}{1 + 2x \left(-\frac{x}{3}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{2x + \frac{x}{3}}{1 - \frac{2x^2}{3}} \right| = \left| \frac{\frac{6x + x}{3}}{\frac{3 - 2x^2}{3}} \right| = \left| \frac{7x}{3 - 2x^2} \right| \]

### Step 4: Evaluate at the intersection points.

Using \( x = 1 \):
\[ \tan \theta = \left| \frac{7 \cdot 1}{3 - 2 \cdot 1^2} \right| = \left| \frac{7}{3 - 2} \right| = \left| \frac{7}{1} \right| = 7 \]
Thus, \( \theta = \tan^{-1}(7) \).

### Conclusion

The angle of intersection of the curves \( y = x^2 \) and \( 6y = 7 - x^2 \) is \( \tan^{-1}(7) \).

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