The curve represented by x = 3 (cos t + sin t), y = 4 (cos t- sin t) is an ellipse whose eccentricity is e, such that `16e^2 + 7` is equal to:

Options

A. 14
B. 12
C. 11
D. 21

Answer

To find the eccentricity \( e \) of the ellipse represented by the parametric equations \( x = 3(\cos t + \sin t) \) and \( y = 4(\cos t - \sin t) \), we will follow these steps:

### Step 1: Express \( \cos t \) and \( \sin t \) in terms of \( x \) and \( y \)

We start with the given equations:
\[ x = 3(\cos t + \sin t) \quad \text{(1)} \]
\[ y = 4(\cos t - \sin t) \quad \text{(2)} \]

From equation (1), we can express \( \cos t + \sin t \):
\[ \cos t + \sin t = \frac{x}{3} \]

From equation (2), we can express \( \cos t - \sin t \):
\[ \cos t - \sin t = \frac{y}{4} \]

### Step 2: Square both expressions

Next, we square both expressions:
\[ (\cos t + \sin t)^2 = \left(\frac{x}{3}\right)^2 \quad \Rightarrow \quad \cos^2 t + 2\cos t \sin t + \sin^2 t = \frac{x^2}{9} \]
\[ (\cos t - \sin t)^2 = \left(\frac{y}{4}\right)^2 \quad \Rightarrow \quad \cos^2 t - 2\cos t \sin t + \sin^2 t = \frac{y^2}{16} \]

### Step 3: Add the squared equations

Now, we add the two squared equations:
\[ \left(\cos^2 t + \sin^2 t + 2\cos t \sin t\right) + \left(\cos^2 t + \sin^2 t - 2\cos t \sin t\right) = \frac{x^2}{9} + \frac{y^2}{16} \]
This simplifies to:
\[ 2(\cos^2 t + \sin^2 t) = \frac{x^2}{9} + \frac{y^2}{16} \]
Since \( \cos^2 t + \sin^2 t = 1 \):
\[ 2 = \frac{x^2}{9} + \frac{y^2}{16} \]

### Step 4: Rearranging to standard form

Rearranging gives us:
\[ \frac{x^2}{9} + \frac{y^2}{16} = 2 \]
Dividing through by 2:
\[ \frac{x^2}{18} + \frac{y^2}{32} = 1 \]

### Step 5: Identify \( a^2 \) and \( b^2 \)

From the standard form of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we identify:
\[ a^2 = 18 \quad \text{and} \quad b^2 = 32 \]

### Step 6: Calculate the eccentricity \( e \)

The eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Calculating \( \frac{b^2}{a^2} \):
\[ \frac{b^2}{a^2} = \frac{32}{18} = \frac{16}{9} \]
Thus,
\[ e = \sqrt{1 - \frac{16}{9}} = \sqrt{\frac{9 - 16}{9}} = \sqrt{\frac{-7}{9}} \quad \text{(This indicates a mistake in the assumption of a and b)} \]

### Step 7: Correcting the values of \( a \) and \( b \)

Since \( b^2 > a^2 \), we should switch \( a \) and \( b \):
\[ a^2 = 32 \quad \text{and} \quad b^2 = 18 \]
Now we recalculate:
\[ e = \sqrt{1 - \frac{18}{32}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 8: Calculate \( 16e^2 + 7 \)

Now we calculate \( 16e^2 + 7 \):
\[ e^2 = \left(\frac{\sqrt{7}}{4}\right)^2 = \frac{7}{16} \]
Thus,
\[ 16e^2 = 16 \times \frac{7}{16} = 7 \]
Finally,
\[ 16e^2 + 7 = 7 + 7 = 14 \]

### Final Answer
The value of \( 16e^2 + 7 \) is \( \boxed{14} \).