Eccentricity of ellipse whose equation is
x=3 (cos t + sin t), y = 4 (cos t - sin t) where t is parameter is

Options

A. `1//2`
B. `1//sqrt3`
C. `sqrt7//4`
D. `2//sqrt3`

Answer

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

### Step 1: Write the equations in terms of \( \cos t \) and \( \sin t \)

From the given parametric equations:
\[ x = 3(\cos t + \sin t) \quad \text{and} \quad y = 4(\cos t - \sin t) \]

### Step 2: Rearrange the equations

We can express \( \cos t \) and \( \sin t \) in terms of \( x \) and \( y \):
\[ \cos t + \sin t = \frac{x}{3} \quad \text{(1)} \]
\[ \cos t - \sin t = \frac{y}{4} \quad \text{(2)} \]

### Step 3: Square both equations

Now, we square both equations (1) and (2):
\[ (\cos t + \sin t)^2 = \left(\frac{x}{3}\right)^2 \quad \Rightarrow \quad \cos^2 t + 2\cos t \sin t + \sin^2 t = \frac{x^2}{9} \]
\[ (\cos t - \sin t)^2 = \left(\frac{y}{4}\right)^2 \quad \Rightarrow \quad \cos^2 t - 2\cos t \sin t + \sin^2 t = \frac{y^2}{16} \]

### Step 4: Use the identity \( \cos^2 t + \sin^2 t = 1 \)

Adding both squared equations:
\[ \left(\cos^2 t + \sin^2 t\right) + \left(\cos^2 t + \sin^2 t\right) = \frac{x^2}{9} + \frac{y^2}{16} \]
This simplifies to:
\[ 2 = \frac{x^2}{9} + \frac{y^2}{16} \]

### Step 5: Rearrange to standard form of the ellipse

Multiply through by 144 (the least common multiple of 9 and 16) to eliminate the denominators:
\[ 288 = 16x^2 + 9y^2 \]
Rearranging gives:
\[ \frac{x^2}{18} + \frac{y^2}{32} = 1 \]

### Step 6: Identify \( a^2 \) and \( b^2 \)

From the standard form of the ellipse \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \):
- \( b^2 = 18 \) (thus \( b = \sqrt{18} = 3\sqrt{2} \))
- \( a^2 = 32 \) (thus \( a = \sqrt{32} = 4\sqrt{2} \))

### Step 7: Calculate \( c \)

Using the relationship \( c^2 = a^2 - b^2 \):
\[ c^2 = 32 - 18 = 14 \quad \Rightarrow \quad c = \sqrt{14} \]

### Step 8: Calculate the eccentricity \( e \)

The eccentricity \( e \) is given by:
\[ e = \frac{c}{a} = \frac{\sqrt{14}}{4\sqrt{2}} = \frac{\sqrt{14}}{8} \]

Thus, the eccentricity of the ellipse is:
\[ e = \frac{\sqrt{7}}{4} \]

### Final Answer:
The eccentricity of the ellipse is \( \frac{\sqrt{7}}{4} \).

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