Find the angle of intersection of the curves :
(i) `y^(2)=4x and x^(2)=4y`

Answer

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

### Step 1: Find Points of Intersection
We need to find the points where the two curves intersect.

1. The first curve is given by:
\[ y^2 = 4x \quad (1) \]
2. The second curve is given by:
\[ x^2 = 4y \quad (2) \]

To find the points of intersection, we can express \( y \) from equation (1) and substitute it into equation (2).

From equation (1):
\[ y = \sqrt{4x} = 2\sqrt{x} \]

Substituting \( y \) into equation (2):
\[ x^2 = 4(2\sqrt{x}) \implies x^2 = 8\sqrt{x} \]

Rearranging gives:
\[ x^2 - 8\sqrt{x} = 0 \]

Let \( \sqrt{x} = t \) (where \( t \geq 0 \)):
\[ t^4 - 8t = 0 \]
Factoring out \( t \):
\[ t(t^3 - 8) = 0 \]
This gives us:
\[ t = 0 \quad \text{or} \quad t^3 = 8 \implies t = 2 \]

Thus, \( \sqrt{x} = 0 \) gives \( x = 0 \) and \( \sqrt{x} = 2 \) gives \( x = 4 \).

Now substituting back to find \( y \):
- For \( x = 0 \):
\[ y = 2\sqrt{0} = 0 \implies (0, 0) \]
- For \( x = 4 \):
\[ y = 2\sqrt{4} = 4 \implies (4, 4) \]

### Step 2: Calculate the Slopes of the Tangents
Next, we need to find the slopes of the tangents to the curves at the points of intersection.

#### For the curve \( y^2 = 4x \):
Differentiating implicitly:
\[ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

At the point \( (4, 4) \):
\[ \frac{dy}{dx} = \frac{2}{4} = \frac{1}{2} \quad (m_1) \]

#### For the curve \( x^2 = 4y \):
Differentiating implicitly:
\[ 2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} \]

At the point \( (4, 4) \):
\[ \frac{dy}{dx} = \frac{4}{2} = 2 \quad (m_2) \]

### Step 3: Find the Angle Between the Curves
The angle \( \theta \) between the two curves can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting \( m_1 = \frac{1}{2} \) and \( m_2 = 2 \):
\[ \tan \theta = \left| \frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \cdot 2} \right| = \left| \frac{\frac{1}{2} - \frac{4}{2}}{1 + 1} \right| = \left| \frac{-\frac{3}{2}}{2} \right| = \frac{3}{4} \]

### Step 4: Calculate the Angle
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Conclusion
The angles of intersection of the curves \( y^2 = 4x \) and \( x^2 = 4y \) are:
1. At the point \( (0, 0) \): \( 90^\circ \)
2. At the point \( (4, 4) \): \( \tan^{-1}\left(\frac{3}{4}\right) \)