The distance from the foci of `P (x_(1), y_(1))` on the ellipse `x^2/9+y^2/25=1` are

Options

A. `4+-(5)/(4) y_(1)`
B. `5+-(4)/(5) x_(1)`
C. `5+-(5)/(4) y_(1)`
D. none of these

Answer

To solve the problem of finding the distance from the foci of the point \( P(x_1, y_1) \) on the ellipse given by the equation

\[ \frac{x^2}{9} + \frac{y^2}{25} = 1, \]

we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given equation of the ellipse can be rewritten in the standard form:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

where \( a^2 = 9 \) and \( b^2 = 25 \). Thus, we have:

- \( a = \sqrt{9} = 3 \)
- \( b = \sqrt{25} = 5 \)

### Step 2: Determine the orientation of the ellipse
Since \( b > a \), the major axis of the ellipse is along the y-axis. Therefore, the foci of the ellipse are located at \( (0, \pm c) \), where \( c \) is given by:

\[ c = \sqrt{b^2 - a^2} = \sqrt{25 - 9} = \sqrt{16} = 4. \]

Thus, the coordinates of the foci are \( (0, 4) \) and \( (0, -4) \).

### Step 3: Calculate the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse is defined as:

\[ e = \frac{c}{b} = \frac{4}{5}. \]

### Step 4: Find the distance from the point \( P(x_1, y_1) \) to the foci
The distance \( D \) from the point \( P(x_1, y_1) \) to the foci can be calculated using the formula for the distance from a point to the foci of the ellipse. Since the major axis is along the y-axis, the distance from the point \( P(x_1, y_1) \) to the foci is given by:

\[ D = b - e \cdot |y_1|, \]

where \( b = 5 \) and \( e = \frac{4}{5} \).

### Step 5: Substitute the values into the formula
Substituting the values we have:

\[ D = 5 - \left(\frac{4}{5}\right) |y_1| = 5 - \frac{4}{5} |y_1|. \]

### Final Result
Thus, the distance from the foci of the point \( P(x_1, y_1) \) on the ellipse is:

\[ D = 5 - \frac{4}{5} |y_1|. \]