If `e` and `e'` be the eccentricities of two conics `S=0` and `S'=0` and if `e^(2)+e'^(2)=3`, then both `S` and `S'` can be

Options

A. hyperbola
B. ellipse
C. parabola
D. None of these

Answer

To solve the problem, we need to analyze the eccentricities of the conics given the condition \( e^2 + e'^2 = 3 \). We will check each type of conic section: parabola, ellipse, and hyperbola.

### Step 1: Analyze the Parabola
- The eccentricity \( e \) of a parabola is always equal to 1.
- Therefore, if both conics are parabolas, we have:
\[ e = 1 \quad \text{and} \quad e' = 1 \]
- Now, substituting these values into the equation:
\[ e^2 + e'^2 = 1^2 + 1^2 = 1 + 1 = 2 \]
- This does not satisfy the condition \( e^2 + e'^2 = 3 \).

**Conclusion**: Both conics cannot be parabolas.

### Step 2: Analyze the Ellipse
- The eccentricity \( e \) of an ellipse is always less than 1, i.e., \( e < 1 \).
- Therefore, if both conics are ellipses, we have:
\[ e < 1 \quad \text{and} \quad e' < 1 \]
- Squaring these values gives:
\[ e^2 < 1 \quad \text{and} \quad e'^2 < 1 \]
- Thus, summing these:
\[ e^2 + e'^2 < 1 + 1 = 2 \]
- This also does not satisfy the condition \( e^2 + e'^2 = 3 \).

**Conclusion**: Both conics cannot be ellipses.

### Step 3: Analyze the Hyperbola
- The eccentricity \( e \) of a hyperbola is always greater than 1, i.e., \( e > 1 \).
- Therefore, if both conics are hyperbolas, we have:
\[ e > 1 \quad \text{and} \quad e' > 1 \]
- Squaring these values gives:
\[ e^2 > 1 \quad \text{and} \quad e'^2 > 1 \]
- Thus, summing these:
\[ e^2 + e'^2 > 1 + 1 = 2 \]
- This means \( e^2 + e'^2 \) can be equal to 3 (since it is greater than 2).

**Conclusion**: Both conics can be hyperbolas.

### Final Answer
Both \( S \) and \( S' \) can be hyperbolas.

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