Question

The eccentricity of the hyperbola \( \frac{(2x-6)^2}{2} - \frac{(4y+7)^2}{16} = 1 \) is

• A. \( \sqrt{5} \)
• B. \( \frac{\sqrt{5}}{2} \)
• C. \( \sqrt{3} \)
• D. \( \sqrt{10} \)
• E. \( \frac{\sqrt{3}}{2} \)

Hint:

Always simplify coefficients properly before identifying \(a^2\) and \(b^2\).

Answer 1

The Correct Option is C

Concept: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Step 1: Convert to standard form.
\[ \frac{(2(x-3))^2}{2} - \frac{(4(y+\frac{7}{4}))^2}{16} = 1 \] \[ \Rightarrow \frac{4(x-3)^2}{2} - \frac{16(y+\frac{7}{4})^2}{16} = 1 \] \[ \Rightarrow \frac{(x-3)^2}{\frac{1}{2}} - \frac{(y+\frac{7}{4})^2}{1} = 1 \] Thus, \[ a^2 = \frac{1}{2}, \quad b^2 = 1 \]

Step 2: Find eccentricity.
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{1/2}} = \sqrt{1 + 2} = \sqrt{3} \]