Question

The distance between the foci of the ellipse \(\frac{(x + 2)^{2{9} + \frac{(y - 1)^{2}}{4} = 1\) is}

• A. \(\sqrt{5}\)
• B. \(2\sqrt{5}\)
• C. \(3\sqrt{5}\)
• D. \(9\sqrt{5}\)
• E. \(7\sqrt{5}\)

Hint:

You don't always need to calculate eccentricity (\(e\)) separately. The identity \(c^2 = a^2 - b^2\) is the fastest way to find the focal distance \(c\), where the distance between foci is simply \(2c\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The distance between the foci of an ellipse is given by the formula \(2ae\), where \(a\) is the semi-major axis and \(e\) is the eccentricity. The center of the ellipse being shifted to \((-2, 1)\) does not change the internal dimensions like the distance between the foci.

Step 2: Key Formula or Approach:
1. Standard form: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
2. Distance between foci = \(2c\), where \(c^2 = a^2 - b^2\).

Step 3: Detailed Explanation:
1. Compare the given equation to the standard form: \(a^2 = 9\) and \(b^2 = 4\).
2. Since \(a^2 > b^2\), the major axis is horizontal.
3. Calculate the distance of each focus from the center (\(c\)): \[ c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5} \]
4. The total distance between the two foci is \(2c\): \[ \text{Distance} = 2 \times \sqrt{5} = 2\sqrt{5} \]

Step 4: Final Answer
The distance between the foci of the ellipse is \(2\sqrt{5}\).