if ` theta ` is a parameter then `x=a ( sin theta + cos theta),`
`y=b( sin theta - cos theta )` respresents

Options

A. an ellipse
B. a circle
C. a pair of stright lines
D. a hyperbola

Answer

To solve the problem, we need to analyze the given equations for \( x \) and \( y \) in terms of the parameter \( \theta \).

### Step-by-Step Solution:

1. **Given Equations**:
\[ x = a(\sin \theta + \cos \theta) \]
\[ y = b(\sin \theta - \cos \theta) \]

2. **Rearranging the Equations**:
We can express \( \sin \theta \) and \( \cos \theta \) in terms of \( x \) and \( y \):
\[ \frac{x}{a} = \sin \theta + \cos \theta \]
\[ \frac{y}{b} = \sin \theta - \cos \theta \]

3. **Squaring Both Sides**:
Square both equations:
\[ \left(\frac{x}{a}\right)^2 = (\sin \theta + \cos \theta)^2 \]
\[ \left(\frac{y}{b}\right)^2 = (\sin \theta - \cos \theta)^2 \]

4. **Expanding the Squares**:
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ \left(\frac{x}{a}\right)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta \]
\[ \left(\frac{y}{b}\right)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \]

5. **Using the Pythagorean Identity**:
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{x}{a}\right)^2 = 1 + 2\sin \theta \cos \theta \]
\[ \left(\frac{y}{b}\right)^2 = 1 - 2\sin \theta \cos \theta \]

6. **Adding the Two Equations**:
Now, add the two squared equations:
\[ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = (1 + 2\sin \theta \cos \theta) + (1 - 2\sin \theta \cos \theta) \]
This simplifies to:
\[ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 2 \]

7. **Rearranging the Equation**:
Dividing through by 2 gives:
\[ \frac{x^2}{2a^2} + \frac{y^2}{2b^2} = 1 \]

8. **Conclusion**:
The final equation represents an ellipse in the standard form:
\[ \frac{x^2}{(a/\sqrt{2})^2} + \frac{y^2}{(b/\sqrt{2})^2} = 1 \]
Therefore, the locus of the point \( (x, y) \) is an ellipse.

### Final Answer:
The given equations represent an ellipse.