The eccentricity of the ellipse `25x^(2) + 16y^(2) - 150 x - 175 = 0` is

Options

A. `(2)/(5)`
B. `(2)/(3)`
C. `(4)/(5) `
D. `(3)/(5)`

Explanation

To find the eccentricity of the ellipse given by the equation \( 25x^2 + 16y^2 - 150x - 175 = 0 \), we will follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ 25x^2 + 16y^2 - 150x - 175 = 0 \]
We can rearrange it to isolate the terms involving \(x\) and \(y\):
\[ 25x^2 - 150x + 16y^2 = 175 \]

### Step 2: Completing the square for the \(x\) terms
Next, we complete the square for the \(x\) terms:
\[ 25(x^2 - 6x) + 16y^2 = 175 \]
To complete the square for \(x^2 - 6x\), we take half of the coefficient of \(x\) (which is -6), square it, and add it inside the parentheses:
\[ 25\left((x - 3)^2 - 9\right) + 16y^2 = 175 \]
This simplifies to:
\[ 25(x - 3)^2 - 225 + 16y^2 = 175 \]
Adding 225 to both sides gives:
\[ 25(x - 3)^2 + 16y^2 = 400 \]

### Step 3: Dividing by 400 to get the standard form
Now we divide the entire equation by 400 to convert it into the standard form of an ellipse:
\[ \frac{25(x - 3)^2}{400} + \frac{16y^2}{400} = 1 \]
This simplifies to:
\[ \frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1 \]

### 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 can identify:
- \(a^2 = 16\) (which means \(a = 4\))
- \(b^2 = 25\) (which means \(b = 5\))

### Step 5: Calculating the eccentricity
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 \(b^2\):
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{1 - 0.64} = \sqrt{0.36} = \frac{3}{5} \]

### Final Answer
Therefore, the eccentricity of the ellipse is:
\[ \frac{3}{5} \]

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