Let e and l are the eccentricity and length of the lactus rectum respectively of the conic described parametrically by `x=t^(2)+t+1, y=t^(2)-t+1`, then the value of `(e )/(l^(2))` is equal to

Answer

To solve the problem, we need to find the eccentricity \( e \) and the length of the latus rectum \( l \) of the conic described by the parametric equations \( x = t^2 + t + 1 \) and \( y = t^2 - t + 1 \).

### Step 1: Find \( x + y \) and \( x - y \)

Given:
\[ x = t^2 + t + 1 \]
\[ y = t^2 - t + 1 \]

Calculating \( x + y \):
\[ x + y = (t^2 + t + 1) + (t^2 - t + 1) = 2t^2 + 2 \]
Thus,
\[ x + y = 2(t^2 + 1) \]

Calculating \( x - y \):
\[ x - y = (t^2 + t + 1) - (t^2 - t + 1) = 2t \]

### Step 2: Express \( t \) in terms of \( x \) and \( y \)

From \( x - y = 2t \):
\[ t = \frac{x - y}{2} \]

From \( x + y = 2(t^2 + 1) \):
\[ \frac{x + y}{2} = t^2 + 1 \implies t^2 = \frac{x + y}{2} - 1 \]

### Step 3: Substitute \( t \) into the equation

Substituting \( t = \frac{x - y}{2} \) into \( t^2 \):
\[ \left(\frac{x - y}{2}\right)^2 = \frac{x + y}{2} - 1 \]

Expanding:
\[ \frac{(x - y)^2}{4} = \frac{x + y}{2} - 1 \]

### Step 4: Rearranging the equation

Multiply through by 4 to eliminate the fraction:
\[ (x - y)^2 = 2(x + y) - 4 \]

Rearranging gives:
\[ (x - y)^2 = 2x + 2y - 4 \]

### Step 5: Identify the conic section

This equation can be rearranged to the form:
\[ (x - y)^2 = 2(x + y - 2) \]

This is a parabola in the form \( y^2 = 4ax \), where the coefficient of \( x \) gives us the parameter \( a \).

### Step 6: Find eccentricity \( e \) and length of latus rectum \( l \)

For a parabola:
- The eccentricity \( e = 1 \)
- The length of the latus rectum \( l = 4a \)

From our equation, we can see that \( 4a = 2 \), thus:
\[ l = 2 \]

### Step 7: Calculate \( \frac{e}{l^2} \)

Now we can calculate:
\[ \frac{e}{l^2} = \frac{1}{(2)^2} = \frac{1}{4} \]

### Final Answer
The value of \( \frac{e}{l^2} \) is \( 0.25 \).