The eccentricity of the conic `x=3((1-t^(2))/(1+t^(2))) and y=(2t)/(1+t^(2))` is

Options

A. `(2)/(3)`
B. `(2sqrt(2))/(3)`
C. `(3)/(2)`
D. `(3)/(2sqrt(2))`

Answer

To find the eccentricity of the conic given by the parametric equations \( x = 3 \frac{1 - t^2}{1 + t^2} \) and \( y = \frac{2t}{1 + t^2} \), we can follow these steps:

### Step 1: Rewrite the equations
We start with the equations:
\[ x = 3 \frac{1 - t^2}{1 + t^2} \]
\[ y = \frac{2t}{1 + t^2} \]

### Step 2: Substitute \( t = \tan\left(\frac{\theta}{2}\right) \)
To simplify the equations, we can use the Weierstrass substitution, where we let \( t = \tan\left(\frac{\theta}{2}\right) \).

### Step 3: Transform the equations
Using the substitution \( t = \tan\left(\frac{\theta}{2}\right) \), we have:
\[ 1 - t^2 = 1 - \tan^2\left(\frac{\theta}{2}\right) = \frac{\cos\theta}{\cos^2\left(\frac{\theta}{2}\right)} \]
\[ 1 + t^2 = 1 + \tan^2\left(\frac{\theta}{2}\right) = \frac{1}{\cos^2\left(\frac{\theta}{2}\right)} \]
Thus, we can rewrite:
\[ x = 3 \frac{1 - t^2}{1 + t^2} = 3 \cos\theta \]
\[ y = \frac{2t}{1 + t^2} = 2 \sin\theta \]

### Step 4: Express in terms of sine and cosine
Now we have:
\[ \frac{x}{3} = \cos\theta \]
\[ \frac{y}{2} = \sin\theta \]

### Step 5: Use the Pythagorean identity
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ \left(\frac{y}{2}\right)^2 + \left(\frac{x}{3}\right)^2 = 1 \]
This simplifies to:
\[ \frac{y^2}{4} + \frac{x^2}{9} = 1 \]

### Step 6: Identify the standard form of the ellipse
This equation represents an ellipse in standard form:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Here, \( a^2 = 9 \) and \( b^2 = 4 \).

### Step 7: Calculate the eccentricity
The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{9 - 4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Final Answer
Thus, the eccentricity of the conic is:
\[ \frac{\sqrt{5}}{3} \]