If `e and e'` are the eccentricities of the hyperbola `x^2/a^2-y^2/b^2=1 and y^2/b^2-x^2/a^2=1,` then the point `(1/e,1/(e'))` lies on the circle (A) `x^2+y^2=1` (B) `x^2+y^2=2` (C) `x^2+y^2=3` (D) `x^2+y^2=4`

Options

A. `x^(2)+y^(2)=1`
B. `x^(2)+y^(2)=2`
C. `x^(2)+y^(2)=3`
D. `x^(2)+y^(2)=4`

Explanation

To solve the problem, we need to find the eccentricities \( e \) and \( e' \) of the hyperbolas given by the equations \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). Then, we will determine if the point \( \left( \frac{1}{e}, \frac{1}{e'} \right) \) lies on any of the given circles.

### Step 1: Find the eccentricity \( e \) of the first hyperbola
The eccentricity \( e \) of a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}} \]

### Step 2: Find the eccentricity \( e' \) of the second hyperbola
The second hyperbola is given by \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). This can be rewritten as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \]
This is the equation of the conjugate hyperbola. The eccentricity \( e' \) of the conjugate hyperbola is given by:
\[ e' = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{\frac{b^2 + a^2}{b^2}} \]

### Step 3: Calculate \( \frac{1}{e} \) and \( \frac{1}{e'} \)
Now, we can find \( \frac{1}{e} \) and \( \frac{1}{e'} \):
\[ \frac{1}{e} = \frac{1}{\sqrt{1 + \frac{b^2}{a^2}}} = \frac{a}{\sqrt{a^2 + b^2}} \]
\[ \frac{1}{e'} = \frac{1}{\sqrt{1 + \frac{a^2}{b^2}}} = \frac{b}{\sqrt{a^2 + b^2}} \]

### Step 4: Check if the point \( \left( \frac{1}{e}, \frac{1}{e'} \right) \) lies on a circle
Now we need to check if the point \( \left( \frac{1}{e}, \frac{1}{e'} \right) = \left( \frac{a}{\sqrt{a^2 + b^2}}, \frac{b}{\sqrt{a^2 + b^2}} \right) \) lies on any of the circles given by the equations.

Calculate \( \left( \frac{1}{e} \right)^2 + \left( \frac{1}{e'} \right)^2 \):
\[ \left( \frac{1}{e} \right)^2 + \left( \frac{1}{e'} \right)^2 = \left( \frac{a}{\sqrt{a^2 + b^2}} \right)^2 + \left( \frac{b}{\sqrt{a^2 + b^2}} \right)^2 \]
\[ = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1 \]

### Conclusion
Since \( \left( \frac{1}{e}, \frac{1}{e'} \right) \) satisfies the equation \( x^2 + y^2 = 1 \), we conclude that the point lies on the circle described by this equation.

Thus, the correct answer is:
**(A) \( x^2 + y^2 = 1 \)**