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

Step 1: Understanding the Concept:
To find the eccentricity of a hyperbola, we must first convert its equation to the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). From this, we can identify \(a^2\) and \(b^2\) and use the formula relating them to eccentricity `e`.
Step 2: Key Formula or Approach:
1. Standard form of a horizontal hyperbola: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). 2. Relationship for eccentricity: \(b^2 = a^2(e^2 - 1)\).
Step 3: Detailed Explanation:
The given equation is: \[ \frac{(2x-6)^2}{2} - \frac{(4y+7)^2}{16} = 1 \] We need to factor out the coefficients of `x` and `y` to get it into standard form. For the first term: \[ \frac{(2(x-3))^2}{2} = \frac{4(x-3)^2}{2} = \frac{2(x-3)^2}{1} = \frac{(x-3)^2}{1/2} \] For the second term: \[ \frac{(4(y+7/4))^2}{16} = \frac{16(y+7/4)^2}{16} = \frac{(y+7/4)^2}{1} \] Now, substitute these back into the equation: \[ \frac{(x-3)^2}{1/2} - \frac{(y+7/4)^2}{1} = 1 \] By comparing with the standard form, we can identify:
\(a^2 = \frac{1}{2}\)
\(b^2 = 1\)
Now, use the formula for eccentricity: \[ b^2 = a^2(e^2 - 1) \] Substitute the values of \(a^2\) and \(b^2\): \[ 1 = \frac{1}{2}(e^2 - 1) \] Multiply both sides by 2: \[ 2 = e^2 - 1 \] Solve for \(e^2\): \[ e^2 = 3 \] Take the square root to find `e` (eccentricity is always >1 for a hyperbola): \[ e = \sqrt{3} \] Step 4: Final Answer:
The eccentricity of the hyperbola is \(\sqrt{3}\).