x = 4 (1+cos `theta `) and y=3 (1+sin `theta`) are the paramatic equations of

Options

A. `((x-3)^2)/(9) + ((y-4)^2)/(16) = 1`
B. `((x+4)^2)/(16) + ((y+3)^2)/(9) = 1`
C. `((x -4)^2)/(16) - ((y-3)^2)/(9) = 1`
D. `((x - 4)^2)/(16) + ((y - 3)^2)/(9) = 1 `

Answer

To solve the problem, we start with the given parametric equations:

1. \( x = 4(1 + \cos \theta) \)
2. \( y = 3(1 + \sin \theta) \)

### Step-by-step Solution:

**Step 1: Express \(\cos \theta\) in terms of \(x\)**

From the equation \( x = 4(1 + \cos \theta) \), we can isolate \(\cos \theta\):

\[ \frac{x}{4} = 1 + \cos \theta \]

Subtracting 1 from both sides gives:

\[ \cos \theta = \frac{x}{4} - 1 \]

**Step 2: Express \(\sin \theta\) in terms of \(y\)**

From the equation \( y = 3(1 + \sin \theta) \), we can isolate \(\sin \theta\):

\[ \frac{y}{3} = 1 + \sin \theta \]

Subtracting 1 from both sides gives:

\[ \sin \theta = \frac{y}{3} - 1 \]

**Step 3: Use the Pythagorean identity**

We know that:

\[ \cos^2 \theta + \sin^2 \theta = 1 \]

Substituting the expressions we found for \(\cos \theta\) and \(\sin \theta\):

\[ \left( \frac{x}{4} - 1 \right)^2 + \left( \frac{y}{3} - 1 \right)^2 = 1 \]

**Step 4: Expand and simplify the equation**

Expanding both squares:

\[ \left( \frac{x}{4} - 1 \right)^2 = \frac{x^2}{16} - \frac{x}{4} + 1 \]
\[ \left( \frac{y}{3} - 1 \right)^2 = \frac{y^2}{9} - \frac{y}{3} + 1 \]

Substituting back into the equation gives:

\[ \frac{x^2}{16} - \frac{x}{4} + 1 + \frac{y^2}{9} - \frac{y}{3} + 1 = 1 \]

Combining like terms:

\[ \frac{x^2}{16} + \frac{y^2}{9} - \frac{x}{4} - \frac{y}{3} + 1 = 1 \]

Subtracting 1 from both sides:

\[ \frac{x^2}{16} + \frac{y^2}{9} - \frac{x}{4} - \frac{y}{3} = 0 \]

**Step 5: Rearranging the equation**

To make it look like the standard form of an ellipse, we rearrange:

\[ \frac{x - 4}{4}^2 + \frac{y - 3}{3}^2 = 1 \]

This can be rewritten as:

\[ \frac{(x - 4)^2}{16} + \frac{(y - 3)^2}{9} = 1 \]

### Conclusion:

Thus, the parametric equations represent the ellipse given by:

\[ \frac{(x - 4)^2}{16} + \frac{(y - 3)^2}{9} = 1 \]

This matches with the third option provided in the question.