Step 1: Understanding the Concept:
The foci are two fixed points on the major axis of an ellipse that define its shape.
The problem requires finding the straight-line distance between these two points based on the given equation of the ellipse.
Step 2: Key Formula or Approach:
The standard equation of an ellipse is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), assuming \(a^2>b^2\).
The lengths of the semi-major and semi-minor axes are \(a\) and \(b\), respectively.
The eccentricity \(e\) is related to \(a\) and \(b\) by the formula \(b^2 = a^2(1 - e^2)\).
The distance from the center to each focus is \(ae\), so the total distance between the two foci is \(2ae\).
Step 3: Detailed Explanation:
The given equation of the ellipse is:
\[ \frac{(x+2)^2}{9} + \frac{(y-1)^2}{4} = 1 \]
By comparing this to the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we can deduce the squared lengths of the semi-axes:
The denominator of the \(x\)-term gives \(a^2 = 9\), so \(a = 3\).
The denominator of the \(y\)-term gives \(b^2 = 4\), so \(b = 2\).
Since \(a>b\), the major axis is horizontal.
Next, we calculate the eccentricity \(e\) using the defining relationship for an ellipse:
\[ b^2 = a^2(1 - e^2) \]
Substitute the values of \(a^2\) and \(b^2\):
\[ 4 = 9(1 - e^2) \]
Divide by \(9\):
\[ \frac{4}{9} = 1 - e^2 \]
Rearrange to solve for \(e^2\):
\[ e^2 = 1 - \frac{4}{9} \]
\[ e^2 = \frac{9 - 4}{9} = \frac{5}{9} \]
Taking the square root (eccentricity is positive):
\[ e = \frac{\sqrt{5}}{3} \]
The formula for the distance between the foci is \(2ae\).
Substitute the values of \(a\) and \(e\) into this formula:
\[ \text{Distance} = 2 \cdot 3 \cdot \left(\frac{\sqrt{5}}{3}\right) \]
The \(3\) in the numerator and denominator cancel out:
\[ \text{Distance} = 2\sqrt{5} \]
Step 4: Final Answer:
The distance between the foci is \(2\sqrt{5}\).