Let the eccentricity of the hyperbola with the principal axes along the coordinate axes and passing through (3, 0) and `(3sqrt2,2)` is e, then the value of `((e^(2)+1)/(e^(2)-1))` is equal to

Answer

To solve the problem, we need to find the value of \(\frac{e^2 + 1}{e^2 - 1}\) for a hyperbola that passes through the points (3, 0) and \((3\sqrt{2}, 2)\).

### Step 1: Write the standard equation of the hyperbola
The standard form of a hyperbola with its principal axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the first point (3, 0)
Substituting the point (3, 0) into the hyperbola equation:
\[ \frac{3^2}{a^2} - \frac{0^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} = 1 \implies a^2 = 9 \]

### Step 3: Substitute the second point \((3\sqrt{2}, 2)\)
Now, substitute the second point \((3\sqrt{2}, 2)\) into the hyperbola equation:
\[ \frac{(3\sqrt{2})^2}{a^2} - \frac{2^2}{b^2} = 1 \]
Substituting \(a^2 = 9\):
\[ \frac{18}{9} - \frac{4}{b^2} = 1 \]
This simplifies to:
\[ 2 - \frac{4}{b^2} = 1 \]
Rearranging gives:
\[ 2 - 1 = \frac{4}{b^2} \implies 1 = \frac{4}{b^2} \implies b^2 = 4 \]

### Step 4: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{9 + 4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 5: Calculate \(e^2\)
Now, we calculate \(e^2\):
\[ e^2 = \left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9} \]

### Step 6: Substitute \(e^2\) into the expression
Now we substitute \(e^2\) into the expression \(\frac{e^2 + 1}{e^2 - 1}\):
\[ \frac{e^2 + 1}{e^2 - 1} = \frac{\frac{13}{9} + 1}{\frac{13}{9} - 1} \]
This simplifies to:
\[ \frac{\frac{13}{9} + \frac{9}{9}}{\frac{13}{9} - \frac{9}{9}} = \frac{\frac{22}{9}}{\frac{4}{9}} = \frac{22}{4} = \frac{11}{2} \]

### Final Answer
Thus, the value of \(\frac{e^2 + 1}{e^2 - 1}\) is:
\[ \frac{11}{2} \text{ or } 5.5 \]

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