The curve with parametric equations
`x=1 +4cos theta , y= 2 +3 sin theta .` is

Options

A. an ellipse
B. a parabola
C. a hyperbola
D. a circle

Explanation

To determine the type of curve represented by the parametric equations \( x = 1 + 4 \cos \theta \) and \( y = 2 + 3 \sin \theta \), we can follow these steps:

### Step 1: Rewrite the parametric equations
We start with the given parametric equations:
\[ x = 1 + 4 \cos \theta \]
\[ y = 2 + 3 \sin \theta \]

### Step 2: Isolate the trigonometric functions
From the first equation, we can isolate \( \cos \theta \):
\[ \cos \theta = \frac{x - 1}{4} \]
From the second equation, we can isolate \( \sin \theta \):
\[ \sin \theta = \frac{y - 2}{3} \]

### Step 3: Use the Pythagorean identity
We know the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting the expressions we found for \( \sin \theta \) and \( \cos \theta \):
\[ \left(\frac{y - 2}{3}\right)^2 + \left(\frac{x - 1}{4}\right)^2 = 1 \]

### Step 4: Square both sides
Squaring both sides gives us:
\[ \frac{(y - 2)^2}{9} + \frac{(x - 1)^2}{16} = 1 \]

### Step 5: Rearranging to standard form
This equation can be rearranged to match the standard form of an ellipse:
\[ \frac{(x - 1)^2}{16} + \frac{(y - 2)^2}{9} = 1 \]

### Step 6: Identify the type of curve
The equation \(\frac{(x - 1)^2}{16} + \frac{(y - 2)^2}{9} = 1\) is in the standard form of an ellipse, which is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \((h, k)\) is the center of the ellipse, and \(a\) and \(b\) are the semi-major and semi-minor axes respectively.

### Conclusion
Thus, the curve represented by the given parametric equations is an **ellipse**.

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