Foci of ellipse `((x-1)^(2))/(4)+((y-2)^(2))/(9)=1` is/are

Options

A. `(1,2) +sqrt(5))`
B. `(1,2- sqrt(3))`
C. `(1+(2sqrt(5))/(3),2)`
D. `(1-(2 sqrt(5))/(3),2)`

Answer

To find the foci of the ellipse given by the equation

\[ \frac{(x-1)^2}{4} + \frac{(y-2)^2}{9} = 1, \]

we will follow these steps:

### Step 1: Identify the standard form of the ellipse
The standard form of an ellipse is given by

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \]

where \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

From the given equation, we can identify:
- \(h = 1\)
- \(k = 2\)
- \(a^2 = 4 \Rightarrow a = 2\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 2: Determine the orientation of the ellipse
Since \(b^2 > a^2\) (i.e., \(9 > 4\)), the major axis is vertical.

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of an ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{a^2}{b^2}}. \]

Substituting the values of \(a^2\) and \(b^2\):

\[ e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}. \]

### Step 4: Find the coordinates of the foci
For an ellipse with a vertical major axis, the foci are located at:

\[ (h, k \pm be). \]

Substituting the values of \(h\), \(k\), \(b\), and \(e\):

- \(b = 3\)
- \(e = \frac{\sqrt{5}}{3}\)

Calculating \(be\):

\[ be = 3 \cdot \frac{\sqrt{5}}{3} = \sqrt{5}. \]

Thus, the coordinates of the foci are:

\[ (1, 2 + \sqrt{5}) \quad \text{and} \quad (1, 2 - \sqrt{5}). \]

### Final Answer
The foci of the ellipse are:

\[ (1, 2 + \sqrt{5}) \quad \text{and} \quad (1, 2 - \sqrt{5}). \]

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