If `(3)/(2+e^(itheta)) =ax + iby`, then the locous of `P(x,y)` will represent

Options

A. (a) ellipse of a =1,b=2
B. (b) circle if a=b=1
C. (c) pair of straight line if a=1,b=0
D. (d) None of these

Explanation

To solve the problem, we start with the equation given:

\[ \frac{3}{2 + e^{i\theta}} = ax + iby \]

### Step 1: Rearranging the equation

First, we can rearrange the equation to isolate \( e^{i\theta} \):

\[ 3 = (2 + e^{i\theta})(ax + iby) \]

### Step 2: Expanding the right-hand side

Now, we expand the right-hand side:

\[ 3 = 2(ax + iby) + e^{i\theta}(ax + iby) \]

This gives us:

\[ 3 = 2ax + 2iby + ax e^{i\theta} + iby e^{i\theta} \]

### Step 3: Separating real and imaginary parts

Next, we can separate the real and imaginary parts. The real part is \( 2ax \) and the imaginary part is \( 2by \). Therefore, we can write:

\[ 3 = 2ax + (2by + ax e^{i\theta} + iby e^{i\theta}) \]

### Step 4: Taking magnitudes

We know that the magnitude of \( e^{i\theta} \) is 1, so we take the magnitude of both sides:

\[ \left| \frac{3}{2 + e^{i\theta}} \right| = 1 \]

### Step 5: Finding the magnitude of the left side

Calculating the magnitude on the left side:

\[ |3| = 3 \quad \text{and} \quad |2 + e^{i\theta}| = |2 + \cos(\theta) + i\sin(\theta)| \]

Using the formula for the magnitude of a complex number:

\[ |2 + e^{i\theta}| = \sqrt{(2 + \cos(\theta))^2 + \sin^2(\theta)} = \sqrt{(2 + \cos(\theta))^2 + (1 - \cos^2(\theta))} \]

### Step 6: Simplifying the expression

Now, simplifying the expression:

\[ = \sqrt{(2 + \cos(\theta))^2 + 1 - \cos^2(\theta)} = \sqrt{(2 + \cos(\theta))^2 + 1 - \cos^2(\theta)} \]

### Step 7: Setting up the equation

Now we have:

\[ 3 = |2 + e^{i\theta}| \]

### Step 8: Analyzing the locus

To analyze the locus of \( P(x, y) \), we can substitute values for \( a \) and \( b \) to see what shapes we get:

1. If \( a = 1 \) and \( b = 2 \), we find that the locus represents an ellipse.
2. If \( a = 1 \) and \( b = 1 \), the locus represents a circle.
3. If \( a = 1 \) and \( b = 0 \), the locus represents a pair of straight lines.

### Conclusion

Thus, the locus of \( P(x, y) \) will represent different shapes based on the values of \( a \) and \( b \).

---