If `e_(1)` be the eccentricity of a hyperbola and `e_(2)` be the eccentricity of its conjugate, then show that the point `(1/e_(1) , 1/e_(2))` lies on the circle `x^(2) + y^(2) = 1`.

Answer

To show that the point \((\frac{1}{e_1}, \frac{1}{e_2})\) lies on the circle \(x^2 + y^2 = 1\), we will follow these steps:

### Step 1: Define the eccentricities
The eccentricity \(e_1\) of a hyperbola with the standard equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is given by:
\[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} \]

The eccentricity \(e_2\) of its conjugate hyperbola, which has the equation \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\), is given by:
\[ e_2 = \sqrt{1 + \frac{a^2}{b^2}} \]

### Step 2: Find \(\frac{1}{e_1}\) and \(\frac{1}{e_2}\)
Now, we will calculate \(\frac{1}{e_1}\) and \(\frac{1}{e_2}\):
\[ \frac{1}{e_1} = \frac{1}{\sqrt{1 + \frac{b^2}{a^2}}} = \frac{1}{\sqrt{\frac{a^2 + b^2}{a^2}}} = \frac{a}{\sqrt{a^2 + b^2}} \]

\[ \frac{1}{e_2} = \frac{1}{\sqrt{1 + \frac{a^2}{b^2}}} = \frac{1}{\sqrt{\frac{b^2 + a^2}{b^2}}} = \frac{b}{\sqrt{a^2 + b^2}} \]

### Step 3: Substitute into the circle equation
Now, we substitute \(\frac{1}{e_1}\) and \(\frac{1}{e_2}\) into the circle equation \(x^2 + y^2 = 1\):
\[ \left(\frac{1}{e_1}\right)^2 + \left(\frac{1}{e_2}\right)^2 = \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2 \]

Calculating the squares:
\[ = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} \]

### Step 4: Simplify the expression
Combining the fractions:
\[ = \frac{a^2 + b^2}{a^2 + b^2} = 1 \]

### Conclusion
Since we have shown that:
\[ \left(\frac{1}{e_1}\right)^2 + \left(\frac{1}{e_2}\right)^2 = 1 \]
This means the point \(\left(\frac{1}{e_1}, \frac{1}{e_2}\right)\) lies on the circle \(x^2 + y^2 = 1\).