The angle of intersection of the curves `y^2=x` and `x^2=y` is :

Options

A. `tan^(-1)(1/2)`
B. `tan^(-1)(5/2)`
C. `tan^(-1)(3/4)`
D. `tan^(-1)(5/4)`

Answer

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

### Step 1: Find the points of intersection
We have the two equations:
1. \(y^2 = x\) (Equation 1)
2. \(x^2 = y\) (Equation 2)

Substituting \(y = x^2\) from Equation 2 into Equation 1:
\[ (x^2)^2 = x \implies x^4 - x = 0 \]
Factoring out \(x\):
\[ x(x^3 - 1) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^3 - 1 = 0 \implies x = 1 \]

### Step 2: Find corresponding \(y\) values
For \(x = 0\):
\[ y^2 = 0 \implies y = 0 \]
Thus, one point of intersection is \((0, 0)\).

For \(x = 1\):
\[ y^2 = 1 \implies y = 1 \]
Thus, the second point of intersection is \((1, 1)\).

### Step 3: Calculate the derivatives to find slopes
Now we differentiate both equations to find the slopes at the points of intersection.

For Equation 1 (\(y^2 = x\)):
\[ 2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y} \]

For Equation 2 (\(x^2 = y\)):
\[ 2x = \frac{dy}{dx} \implies \frac{dy}{dx} = 2x \]

### Step 4: Evaluate the slopes at the intersection points
1. At the point \((0, 0)\):
- From Equation 1: \(y = 0\) gives \(\frac{dy}{dx} = \frac{1}{2(0)}\) (undefined).
- From Equation 2: \(x = 0\) gives \(\frac{dy}{dx} = 2(0) = 0\).

2. At the point \((1, 1)\):
- From Equation 1: \(y = 1\) gives \(\frac{dy}{dx} = \frac{1}{2(1)} = \frac{1}{2}\).
- From Equation 2: \(x = 1\) gives \(\frac{dy}{dx} = 2(1) = 2\).

Thus, we have:
- Slope \(m_1 = \frac{1}{2}\)
- Slope \(m_2 = 2\)

### Step 5: Calculate the angle of intersection
The formula for the tangent of the angle \(\theta\) between two curves is:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan \theta = \left| \frac{2 - \frac{1}{2}}{1 + \left(\frac{1}{2}\right)(2)} \right| = \left| \frac{2 - 0.5}{1 + 1} \right| = \left| \frac{1.5}{2} \right| = \frac{3}{4} \]

### Step 6: Find the angle \(\theta\)
To find the angle \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Final Answer
The angle of intersection of the curves \(y^2 = x\) and \(x^2 = y\) is:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

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