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: Identify the Conic Section
The given equations are in parametric form. We can recognize that they resemble the parametric equations for an ellipse.

### Step 2: Convert Parametric Equations to Cartesian Form
We can express \( x \) and \( y \) in terms of \( t \):

1. From \( x = 3 \frac{1 - t^2}{1 + t^2} \), we can rearrange to find \( t^2 \):
\[ x(1 + t^2) = 3(1 - t^2) \implies xt^2 + x = 3 - 3t^2 \implies (x + 3)t^2 = 3 - x \implies t^2 = \frac{3 - x}{x + 3} \]

2. From \( y = \frac{2t}{1 + t^2} \), we can express \( t \):
\[ y(1 + t^2) = 2t \implies yt^2 - 2t + y = 0 \]
This is a quadratic in \( t \). Using the quadratic formula, we can solve for \( t \):
\[ t = \frac{2 \pm \sqrt{(2)^2 - 4y(y)}}{2y} = \frac{2 \pm \sqrt{4 - 4y^2}}{2y} = \frac{1 \pm \sqrt{1 - y^2}}{y} \]

### Step 3: Substitute \( t^2 \) into \( y \)
Using the expression for \( t^2 \) in terms of \( x \), we can substitute into the equation for \( y \):
\[ y^2 = \left(\frac{2t}{1 + t^2}\right)^2 = \frac{4t^2}{(1 + t^2)^2} \]
Substitute \( t^2 = \frac{3 - x}{x + 3} \):
\[ y^2 = \frac{4 \cdot \frac{3 - x}{x + 3}}{\left(1 + \frac{3 - x}{x + 3}\right)^2} \]

### Step 4: Identify the Standard Form of the Ellipse
After simplification, we can express the conic in the standard form of an ellipse:
\[ \frac{x^2}{9} + \frac{y^2}{1} = 1 \]
This indicates that the semi-major axis \( a = 3 \) and the semi-minor axis \( b = 1 \).

### Step 5: Calculate the Eccentricity
The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( a = 3 \) and \( b = 1 \):
\[ e = \sqrt{1 - \frac{1^2}{3^2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

### Final Answer
The eccentricity of the conic is \( \frac{2\sqrt{2}}{3} \).

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