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:

For ellipses, always remember: \[ c^2 = a^2 - b^2 \] and the distance between foci is \(2c\).

Answer 1

The Correct Option is B

Concept: The standard form of an ellipse is: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where:
• \( a^2 > b^2 \) for a horizontal ellipse,
• The distance between the foci is \( 2c \),
• \( c = \sqrt{a^2 - b^2} \).

Step 1: Identify \( a^2 \) and \( b^2 \). Given: \[ \frac{(x+2)^2}{9} + \frac{(y-1)^2}{4} = 1 \] Here: \[ a^2 = 9, b^2 = 4 \]

Step 2: Find \( c \). \[ c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5} \]

Step 3: Distance between the foci. \[ \text{Distance} = 2c = 2\sqrt{5} \]