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Let the eccentricity of the hyperbola wi...

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

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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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