Angle of intersection of the following curves
`y^2=16 x,2x^2+y^2=4` is `pi//2`

Answer

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

### Step 1: Find the points of intersection
To find the points where the curves intersect, we need to solve the equations simultaneously.

1. From the first curve, we have:
\[ y^2 = 16x \quad \text{(1)} \]

2. Substitute \( y^2 \) from equation (1) into the second curve:
\[ 2x^2 + 16x = 4 \]
Rearranging gives:
\[ 2x^2 + 16x - 4 = 0 \]
Dividing the entire equation by 2:
\[ x^2 + 8x - 2 = 0 \]

3. Now, we can use the quadratic formula to find \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 + 8}}{2} = \frac{-8 \pm \sqrt{72}}{2} = \frac{-8 \pm 6\sqrt{2}}{2} \]
Simplifying gives:
\[ x = -4 \pm 3\sqrt{2} \]

4. Now, substitute these \( x \) values back into equation (1) to find corresponding \( y \) values:
\[ y^2 = 16(-4 \pm 3\sqrt{2}) \Rightarrow y = \pm 4\sqrt{-1 + \frac{3\sqrt{2}}{2}} \text{ (valid only for } x = -4 + 3\sqrt{2} \text{)} \]

### Step 2: Find the slopes of the tangents at the points of intersection
1. For the first curve \( y^2 = 16x \):
- Differentiate implicitly:
\[ 2y \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y} \]
Let \( m_1 = \frac{8}{y} \).

2. For the second curve \( 2x^2 + y^2 = 4 \):
- Differentiate implicitly:
\[ 4x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{y} \]
Let \( m_2 = -\frac{4x}{y} \).

### Step 3: Find the angle of intersection
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 \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{\frac{8}{y} + \frac{4x}{y}}{1 + \left(\frac{8}{y}\right)\left(-\frac{4x}{y}\right)} \right| = \left| \frac{\frac{8 + 4x}{y}}{1 - \frac{32x}{y^2}} \right| \]

### Step 4: Substitute the values at the intersection points
Using \( y^2 = 16x \) gives \( y^2 - 16x = 0 \). Substitute this into the equation:
\[ \tan \theta = \left| \frac{8 + 4x}{y} \cdot \frac{y^2}{y^2 - 16x} \right| \]
Since \( y^2 - 16x = 0 \), we find that \( \tan \theta \) approaches infinity, indicating that \( \theta = \frac{\pi}{2} \).

### Conclusion
Thus, the angle of intersection of the curves \( y^2 = 16x \) and \( 2x^2 + y^2 = 4 \) is \( \frac{\pi}{2} \).

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