A point `P(x , y)` moves in the xy-plane such that `x=acos^2theta` and `y=2asintheta,` where `theta` is a parameter. The locus of the point `P` is a/an circle (b) aellipse unbounded parabola (d) part of the parabola

Options

A. circle
B. ellipse
C. unbounded parabola
D. part of the parabola

Explanation

To find the locus of the point \( P(x, y) \) given the equations \( x = a \cos^2 \theta \) and \( y = 2a \sin \theta \), we will eliminate the parameter \( \theta \) and derive the relationship between \( x \) and \( y \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ x = a \cos^2 \theta \]
\[ y = 2a \sin \theta \]

2. **Express \( \cos^2 \theta \) and \( \sin \theta \) in terms of \( x \) and \( y \):**
From the first equation, we can express \( \cos^2 \theta \):
\[ \cos^2 \theta = \frac{x}{a} \]
From the second equation, we can express \( \sin \theta \):
\[ \sin \theta = \frac{y}{2a} \]

3. **Use the Pythagorean identity:**
Recall the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute the expressions for \( \sin^2 \theta \) and \( \cos^2 \theta \):
\[ \left(\frac{y}{2a}\right)^2 + \left(\frac{x}{a}\right) = 1 \]

4. **Simplify the equation:**
\[ \frac{y^2}{4a^2} + \frac{x}{a} = 1 \]
Multiply through by \( 4a^2 \) to eliminate the denominators:
\[ y^2 + 4ax = 4a^2 \]

5. **Rearrange the equation:**
\[ y^2 = 4a^2 - 4ax \]
This can be rewritten as:
\[ y^2 = 4a(a - x) \]

6. **Identify the type of conic section:**
The equation \( y^2 = 4a(a - x) \) represents a parabola that opens to the left.

7. **Determine the bounds of the locus:**
Since \( \cos^2 \theta \) ranges from \( 0 \) to \( 1 \), \( x \) will range from \( 0 \) to \( a \). The \( y \) values will range from \( -2a \) to \( 2a \) based on the sine function. Thus, the locus is a part of the parabola.

### Conclusion:
The locus of the point \( P(x, y) \) is a part of the parabola.