The foci of the conic `25x^(2) +16y^(2)-150 x=175` are :

Options

A. `(0,+-3)`
B. `(0,+- 2)`
C. `(3,+- 3)`
D. `(0,+-1)`

Explanation

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

### Step 1: Rearranging the Equation
First, we rearrange the equation to group the \( x \) and \( y \) terms together:
\[ 25x^2 - 150x + 16y^2 = 175 \]

### Step 2: Completing the Square for \( x \)
Next, we complete the square for the \( x \) terms. We factor out 25 from the \( x \) terms:
\[ 25(x^2 - 6x) + 16y^2 = 175 \]
To complete the square for \( x^2 - 6x \), we take half of -6, which is -3, square it to get 9, and rewrite the equation:
\[ 25(x^2 - 6x + 9 - 9) + 16y^2 = 175 \]
This simplifies to:
\[ 25((x - 3)^2 - 9) + 16y^2 = 175 \]
Distributing the 25 gives:
\[ 25(x - 3)^2 - 225 + 16y^2 = 175 \]
Now, we add 225 to both sides:
\[ 25(x - 3)^2 + 16y^2 = 400 \]

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

### Step 4: Identifying Parameters
From the standard form of the ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
we identify:
- \( h = 3 \)
- \( k = 0 \)
- \( a^2 = 16 \) (thus \( a = 4 \))
- \( b^2 = 25 \) (thus \( b = 5 \))

### Step 5: Finding the Eccentricity
The eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values we found:
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 6: Finding the Foci
The foci of an ellipse are located at \( (h \pm ae, k) \). Substituting the values:
\[ Foci = (3 \pm 4 \cdot \frac{3}{5}, 0) = (3 \pm \frac{12}{5}, 0) \]
Calculating the coordinates:
- For the positive focus:
\[ 3 + \frac{12}{5} = \frac{15}{5} + \frac{12}{5} = \frac{27}{5} \]
- For the negative focus:
\[ 3 - \frac{12}{5} = \frac{15}{5} - \frac{12}{5} = \frac{3}{5} \]

Thus, the coordinates of the foci are:
\[ \left( \frac{27}{5}, 0 \right) \text{ and } \left( \frac{3}{5}, 0 \right) \]

### Final Answer
The foci of the conic are \( \left( \frac{27}{5}, 0 \right) \) and \( \left( \frac{3}{5}, 0 \right) \).

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