A particle of mass m movies along a curve `y=x^(2)`. When particle has x-co-ordinate as `1//2` and x-component of velocity as `4 m//s`. Then :-

Options

A. The position coordinate of particle are `(1//2, 1//4)`
B. The velocity of particle will be along the line `4x-4y-1=0`
C. The magnitude of velocity at that instant is `4sqrt(2) m//s`
D. The magnitude of angular momentum of particle about origin at that position is 0.

Explanation

To solve the problem step by step, we will follow the given information and apply the necessary equations.

### Step 1: Determine the position coordinates of the particle.
Given the curve equation \( y = x^2 \) and the x-coordinate \( x = \frac{1}{2} \):

1. Substitute \( x = \frac{1}{2} \) into the equation:
\[ y = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

2. Therefore, the position coordinates of the particle are:
\[ (x, y) = \left(\frac{1}{2}, \frac{1}{4}\right) \]

### Step 2: Find the equation of the tangent line at the point.
To find the equation of the tangent line at the point \( \left(\frac{1}{2}, \frac{1}{4}\right) \):

1. Differentiate \( y = x^2 \) with respect to \( x \):
\[ \frac{dy}{dx} = 2x \]

2. Evaluate the slope at \( x = \frac{1}{2} \):
\[ \text{slope} = 2 \times \frac{1}{2} = 1 \]

3. Use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = 1 \), \( x_1 = \frac{1}{2} \), and \( y_1 = \frac{1}{4} \):
\[ y - \frac{1}{4} = 1\left(x - \frac{1}{2}\right) \]

4. Rearranging gives:
\[ y - \frac{1}{4} = x - \frac{1}{2} \implies 4y - 1 = 4x - 2 \implies 4x - 4y - 1 = 0 \]

### Step 3: Calculate the magnitude of the velocity.
Given that the x-component of velocity \( v_x = 4 \, \text{m/s} \):

1. Find the y-component of velocity \( v_y \):
- Differentiate \( y = x^2 \) with respect to time \( t \):
\[ \frac{dy}{dt} = 2x \frac{dx}{dt} \implies v_y = 2x v_x \]
- Substitute \( x = \frac{1}{2} \) and \( v_x = 4 \):
\[ v_y = 2 \times \frac{1}{2} \times 4 = 4 \, \text{m/s} \]

2. Calculate the magnitude of the velocity \( v \):
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \, \text{m/s} \]

### Final Answers:
1. Position coordinates: \( \left(\frac{1}{2}, \frac{1}{4}\right) \)
2. Equation of the tangent line: \( 4x - 4y - 1 = 0 \)
3. Magnitude of velocity: \( 4\sqrt{2} \, \text{m/s} \)