The foci of the ellipse `9x^(2)+25y^(2)-36x+50y-164=0` are

Options

A. (6,1)(2,-1)
B. (6,-1)(-2,-1)
C. (-6,1)(-4,-1)
D. (6,1)(2,1)

Answer

To find the foci of the ellipse given by the equation \(9x^2 + 25y^2 - 36x + 50y - 164 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 9x^2 + 25y^2 - 36x + 50y - 164 = 0 \]
We rearrange it to group the \(x\) terms and \(y\) terms together:
\[ 9x^2 - 36x + 25y^2 + 50y - 164 = 0 \]

### Step 2: Completing the Square for \(x\) and \(y\)
Next, we complete the square for the \(x\) terms and \(y\) terms.

For \(x\):
\[ 9(x^2 - 4x) = 9((x - 2)^2 - 4) = 9(x - 2)^2 - 36 \]

For \(y\):
\[ 25(y^2 + 2y) = 25((y + 1)^2 - 1) = 25(y + 1)^2 - 25 \]

Now, substituting back into the equation:
\[ 9((x - 2)^2 - 4) + 25((y + 1)^2 - 1) - 164 = 0 \]
This simplifies to:
\[ 9(x - 2)^2 - 36 + 25(y + 1)^2 - 25 - 164 = 0 \]
Combining constants:
\[ 9(x - 2)^2 + 25(y + 1)^2 - 225 = 0 \]
Thus, we have:
\[ 9(x - 2)^2 + 25(y + 1)^2 = 225 \]

### Step 3: Dividing by 225
Divide the entire equation by 225 to get it in standard form:
\[ \frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1 \]

### Step 4: Identifying \(a^2\) and \(b^2\)
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(h = 2\)
- \(k = -1\)
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 5: Finding the Foci
The foci of an ellipse are given by the formula:
\[ (h \pm c, k) \]
where \(c = \sqrt{a^2 - b^2}\).

Calculating \(c\):
\[ c = \sqrt{25 - 9} = \sqrt{16} = 4 \]

Thus, the foci are:
\[ (2 + 4, -1) \quad \text{and} \quad (2 - 4, -1) \]
This gives us:
\[ (6, -1) \quad \text{and} \quad (-2, -1) \]

### Final Answer:
The foci of the ellipse are \((6, -1)\) and \((-2, -1)\).

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