If a,b are eccentricities of a hyperbola and its conjugate hyperbola then `a^(-2)+b^(-2)=`

Options

A. 4
B. 1
C. `a^(2)b^(2)`
D. `a^(-2)b^(2)`

Answer

To solve the problem, we need to find the value of \( a^{-2} + b^{-2} \) where \( a \) and \( b \) are the eccentricities of a hyperbola and its conjugate hyperbola, respectively.

### Step-by-step Solution:

1. **Understanding the Eccentricity of a Hyperbola:**
The eccentricity \( e \) of a hyperbola defined by the equation
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}} \]
Therefore, we can express \( e \) as:
\[ e^2 = 1 + \frac{b^2}{a^2} \quad \text{or} \quad e^2 = \frac{a^2 + b^2}{a^2} \]

2. **Eccentricity of the Conjugate Hyperbola:**
The conjugate hyperbola is defined by the equation
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]
The eccentricity \( e' \) of the conjugate hyperbola is given by:
\[ e' = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{\frac{b^2 + a^2}{b^2}} \]
Thus, we can express \( e' \) as:
\[ e'^2 = 1 + \frac{a^2}{b^2} \quad \text{or} \quad e'^2 = \frac{b^2 + a^2}{b^2} \]

3. **Finding \( a^{-2} + b^{-2} \):**
We need to find \( \frac{1}{e^2} + \frac{1}{e'^2} \):
\[ \frac{1}{e^2} = \frac{a^2}{a^2 + b^2} \quad \text{and} \quad \frac{1}{e'^2} = \frac{b^2}{a^2 + b^2} \]
Adding these two fractions gives:
\[ \frac{1}{e^2} + \frac{1}{e'^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1 \]

4. **Conclusion:**
Therefore, we conclude that:
\[ a^{-2} + b^{-2} = 1 \]

### Final Answer:
\[ a^{-2} + b^{-2} = 1 \]