Angle of intersection of the curves `y=4-x^2` and `y=x^2` is

Options

A. `pi/2`
B. `tan^(-1)(4/3)`
C. `tan^(-1)(4sqrt2)/7`
D. None

Answer

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

### Step 1: Find the points of intersection of the curves.
To find the points of intersection, we set the two equations equal to each other:

\[ 4 - x^2 = x^2 \]

Rearranging gives:

\[ 4 = 2x^2 \implies x^2 = 2 \implies x = \pm \sqrt{2} \]

Now, substituting \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) back into either equation to find the corresponding \( y \)-values:

For \( x = \sqrt{2} \):

\[ y = (\sqrt{2})^2 = 2 \]

For \( x = -\sqrt{2} \):

\[ y = (-\sqrt{2})^2 = 2 \]

Thus, the points of intersection are \( (\sqrt{2}, 2) \) and \( (-\sqrt{2}, 2) \).

### Step 2: Find the slopes of the tangents at the points of intersection.
We need to find the derivatives of both curves to get the slopes of the tangents.

For the first curve \( y = 4 - x^2 \):

\[ \frac{dy}{dx} = -2x \]

For the second curve \( y = x^2 \):

\[ \frac{dy}{dx} = 2x \]

Now, we calculate the slopes at the points of intersection.

At \( x = \sqrt{2} \):

- Slope of the first curve \( M_1 = -2(\sqrt{2}) = -2\sqrt{2} \)
- Slope of the second curve \( M_2 = 2(\sqrt{2}) = 2\sqrt{2} \)

At \( x = -\sqrt{2} \):

- Slope of the first curve \( M_1' = -2(-\sqrt{2}) = 2\sqrt{2} \)
- Slope of the second curve \( M_2' = 2(-\sqrt{2}) = -2\sqrt{2} \)

### Step 3: Use the formula for the angle of intersection.
The angle \( \theta \) between two curves can be found using the formula:

\[ \tan \theta = \left| \frac{M_1 - M_2}{1 + M_1 M_2} \right| \]

Using the slopes at the point \( (\sqrt{2}, 2) \):

\[ \tan \theta = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 + (-2\sqrt{2})(2\sqrt{2})} \right| \]

Calculating the numerator:

\[ -2\sqrt{2} - 2\sqrt{2} = -4\sqrt{2} \]

Calculating the denominator:

\[ 1 + (-4) = -3 \]

Thus, we have:

\[ \tan \theta = \left| \frac{-4\sqrt{2}}{-3} \right| = \frac{4\sqrt{2}}{3} \]

### Step 4: Find the angle \( \theta \).
To find \( \theta \), we take the arctangent:

\[ \theta = \tan^{-1} \left( \frac{4\sqrt{2}}{3} \right) \]

### Final Answer:
The angle of intersection of the curves \( y = 4 - x^2 \) and \( y = x^2 \) is:

\[ \theta = \tan^{-1} \left( \frac{4\sqrt{2}}{3} \right) \]