The coordinate of a particle moving in a plane are given by ` x(t) = a cos (pt) and y(t) = b sin (pt)` where `a,b (lt a)` and `P` are positive constants of appropriate dimensions . Then

Options

A. The path of the particle is an ellipse
B. The velocity and acceleration of the particle are normal to each other at `t=pi//2p`
C. The acceleration of the particle is always directed towards a focus
D. The distance traveled by the particle in time interval `t=0" to "t=pi//2p` is a

Explanation

To solve the problem, we need to analyze the motion of a particle whose coordinates are given by the equations:

1. \( x(t) = a \cos(pt) \)
2. \( y(t) = b \sin(pt) \)

where \( a \) and \( b \) are constants and \( p \) is a positive constant.

### Step 1: Find the trajectory of the particle

To find the trajectory, we need to eliminate the parameter \( t \) from the equations. We can do this by squaring both equations.

- From the equation for \( x(t) \):
\[ x^2 = a^2 \cos^2(pt) \]

- From the equation for \( y(t) \):
\[ y^2 = b^2 \sin^2(pt) \]

### Step 2: Use the Pythagorean identity

We know from trigonometry that:
\[ \cos^2(pt) + \sin^2(pt) = 1 \]

Now, we can express \( \cos^2(pt) \) and \( \sin^2(pt) \) in terms of \( x \) and \( y \):

- Rearranging the equations gives:
\[ \cos^2(pt) = \frac{x^2}{a^2} \]
\[ \sin^2(pt) = \frac{y^2}{b^2} \]

### Step 3: Add the equations

Now, substituting these into the Pythagorean identity:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

This is the equation of an ellipse, where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Conclusion for Part 1

Thus, the path of the particle is an ellipse.

### Step 4: Analyze the velocity and acceleration

Next, we need to find the velocity and acceleration of the particle at \( t = \frac{\pi}{2p} \).

### Step 5: Find the velocity

The velocity components are given by the derivatives of \( x(t) \) and \( y(t) \):

- \( v_x = \frac{dx}{dt} = -a p \sin(pt) \)
- \( v_y = \frac{dy}{dt} = b p \cos(pt) \)

At \( t = \frac{\pi}{2p} \):
- \( v_x = -a p \sin\left(p \cdot \frac{\pi}{2p}\right) = -a p \cdot 1 = -a p \)
- \( v_y = b p \cos\left(p \cdot \frac{\pi}{2p}\right) = b p \cdot 0 = 0 \)

Thus, the velocity vector at \( t = \frac{\pi}{2p} \) is:
\[ \mathbf{v} = (-a p) \hat{i} + 0 \hat{j} \]

### Step 6: Find the acceleration

The acceleration components are given by the second derivatives:

- \( a_x = \frac{d^2x}{dt^2} = -a p^2 \cos(pt) \)
- \( a_y = \frac{d^2y}{dt^2} = -b p^2 \sin(pt) \)

At \( t = \frac{\pi}{2p} \):
- \( a_x = -a p^2 \cos\left(p \cdot \frac{\pi}{2p}\right) = -a p^2 \cdot 0 = 0 \)
- \( a_y = -b p^2 \sin\left(p \cdot \frac{\pi}{2p}\right) = -b p^2 \cdot 1 = -b p^2 \)

Thus, the acceleration vector at \( t = \frac{\pi}{2p} \) is:
\[ \mathbf{a} = 0 \hat{i} + (-b p^2) \hat{j} \]

### Step 7: Check if velocity and acceleration are normal to each other

The velocity vector is \( (-a p) \hat{i} \) and the acceleration vector is \( 0 \hat{i} + (-b p^2) \hat{j} \). Since the dot product of these two vectors is zero, they are indeed normal to each other.

### Conclusion for Part 2

The velocity and acceleration of the particle are normal to each other at \( t = \frac{\pi}{2p} \).

### Step 8: Analyze the direction of acceleration

The acceleration vector points towards the center of the ellipse, indicating that it is directed towards a focus of the ellipse.

### Conclusion for Part 3

The acceleration of the particle is always directed towards a focus.

### Step 9: Distance traveled by the particle

To find the distance traveled by the particle from \( t = 0 \) to \( t = \frac{2\pi}{p} \), we note that this corresponds to one complete cycle of the motion along the ellipse. The distance traveled is the perimeter of the ellipse, which is not simply \( a \) or \( b \).

### Conclusion for Part 4

The distance traveled by the particle in the time interval \( t = 0 \) to \( t = \frac{2\pi}{p} \) is not equal to \( a \).

### Final Summary

1. The path of the particle is an ellipse.
2. The velocity and acceleration are normal to each other at \( t = \frac{\pi}{2p} \).
3. The acceleration is always directed towards a focus.
4. The distance traveled is not equal to \( a \).