The foci of the ellipse `9x^(2)+ 5(y^(2)-10y +25)=45` are

Options

A. (-2,5),(2,5)
B. (0,3),(0,7)
C. (0,1),(0,9)
D. `(pm4,5)`

Answer

To find the foci of the ellipse given by the equation \(9x^2 + 5(y^2 - 10y + 25) = 45\), we will follow these steps:

### Step 1: Rearrange the equation
Start with the original equation:
\[ 9x^2 + 5(y^2 - 10y + 25) = 45 \]
Distributing the 5:
\[ 9x^2 + 5y^2 - 50y + 125 = 45 \]
Now, move 45 to the left side:
\[ 9x^2 + 5y^2 - 50y + 80 = 0 \]

### Step 2: Complete the square for the \(y\) terms
We need to complete the square for the \(y\) terms:
\[ 5(y^2 - 10y) = 5((y - 5)^2 - 25) = 5(y - 5)^2 - 125 \]
Substituting this back into the equation gives:
\[ 9x^2 + 5(y - 5)^2 - 125 + 80 = 0 \]
This simplifies to:
\[ 9x^2 + 5(y - 5)^2 - 45 = 0 \]
Rearranging gives:
\[ 9x^2 + 5(y - 5)^2 = 45 \]

### Step 3: Divide by 45
To put the equation in standard form, divide everything by 45:
\[ \frac{9x^2}{45} + \frac{5(y - 5)^2}{45} = 1 \]
This simplifies to:
\[ \frac{x^2}{5} + \frac{(y - 5)^2}{9} = 1 \]

### Step 4: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(a^2 = 5\) → \(a = \sqrt{5}\)
- \(b^2 = 9\) → \(b = 3\)
- Center \((h, k) = (0, 5)\)

### Step 5: Find the eccentricity \(e\)
The eccentricity \(e\) is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Step 6: Find the foci
The foci of the ellipse are located at:
\[ (h, k \pm be) = (0, 5 \pm 3 \cdot \frac{2}{3}) = (0, 5 \pm 2) \]
Thus, the foci are:
\[ (0, 7) \quad \text{and} \quad (0, 3) \]

### Final Answer
The foci of the ellipse are \((0, 7)\) and \((0, 3)\).

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