The asending order of the eccentricities `l_(1),l_(2),l_(3)` of the conics `12x^(2) + 3y^(2) = 15, x^(2) +2y^(2) = 6, 4(x-1)^(2) + 3(y+2)^(2) = 12`

Options

A. `l_(1),l_(2),l_(3)`
B. `l_(3),l_(2),l_(1)`
C. `l_(3),l_(1),l_(2)`
D. `l_(2),l_(1),l_(3)`

Answer

To find the ascending order of the eccentricities \( l_1, l_2, l_3 \) of the given conics, we need to analyze each equation and determine their types (ellipse, parabola, hyperbola) and calculate their eccentricities.

### Step 1: Analyze the first conic \( 12x^2 + 3y^2 = 15 \)

1. **Rearranging the equation**:
\[ \frac{12x^2}{15} + \frac{3y^2}{15} = 1 \implies \frac{4x^2}{5} + \frac{y^2}{5} = 1 \]

2. **Identifying the conic type**:
This is in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a^2 = \frac{5}{4} \) and \( b^2 = 5 \). Since both \( a^2 \) and \( b^2 \) are positive, this is an ellipse.

3. **Calculating the eccentricity**:
The eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5/4}{5}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
So, \( l_1 = \frac{\sqrt{3}}{2} \).

### Step 2: Analyze the second conic \( x^2 + 2y^2 = 6 \)

1. **Rearranging the equation**:
\[ \frac{x^2}{6} + \frac{2y^2}{6} = 1 \implies \frac{x^2}{6} + \frac{y^2}{3} = 1 \]

2. **Identifying the conic type**:
This is also in the form of an ellipse since both denominators are positive.

3. **Calculating the eccentricity**:
Here, \( a^2 = 6 \) and \( b^2 = 3 \):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{6}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
So, \( l_2 = \frac{1}{\sqrt{2}} \).

### Step 3: Analyze the third conic \( 4(x-1)^2 + 3(y+2)^2 = 12 \)

1. **Rearranging the equation**:
\[ \frac{4(x-1)^2}{12} + \frac{3(y+2)^2}{12} = 1 \implies \frac{(x-1)^2}{3} + \frac{(y+2)^2}{4} = 1 \]

2. **Identifying the conic type**:
This is again in the form of an ellipse.

3. **Calculating the eccentricity**:
Here, \( a^2 = 4 \) and \( b^2 = 3 \):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
So, \( l_3 = \frac{1}{2} \).

### Step 4: Compare the eccentricities

Now we have:
- \( l_1 = \frac{\sqrt{3}}{2} \approx 0.866 \)
- \( l_2 = \frac{1}{\sqrt{2}} \approx 0.707 \)
- \( l_3 = \frac{1}{2} = 0.5 \)

### Step 5: Ascending order of eccentricities

Arranging them in ascending order:
\[ l_3 < l_2 < l_1 \implies \frac{1}{2} < \frac{1}{\sqrt{2}} < \frac{\sqrt{3}}{2} \]

### Final Answer
The ascending order of the eccentricities is:
\[ l_3, l_2, l_1 \]