If the ellipse with equation `9x^2+25y^2=225`, then find the eccentricity and foci.

Answer

To find the eccentricity and foci of the ellipse given by the equation \(9x^2 + 25y^2 = 225\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the ellipse:
\[ 9x^2 + 25y^2 = 225 \]
To convert this into standard form, we divide every term by 225:
\[ \frac{9x^2}{225} + \frac{25y^2}{225} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 25\) (which means \(a = \sqrt{25} = 5\))
- \(b^2 = 9\) (which means \(b = \sqrt{9} = 3\))

### Step 3: Determine the eccentricity \(e\)
The formula for the eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{1 - 0.36} = \sqrt{0.64} = \frac{4}{5} \]

### Step 4: Find the foci
The foci of the ellipse are located at \((\pm ae, 0)\). We already have \(a = 5\) and \(e = \frac{4}{5}\):
\[ \text{Foci} = \left(\pm a e, 0\right) = \left(\pm 5 \cdot \frac{4}{5}, 0\right) = \left(\pm 4, 0\right) \]

### Final Answer
- The eccentricity \(e\) of the ellipse is \(\frac{4}{5}\).
- The foci of the ellipse are at the points \((4, 0)\) and \((-4, 0)\).

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