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 pm 5/4 y_1`
B. `5 pm 4/5 x_1`
C. `5 pm 4/5 y_1`
D. none of these

Answer

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

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

we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The standard form of the ellipse is

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

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

\[ a = 3 \quad \text{and} \quad b = 5. \]

### Step 2: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{a^2}{b^2}}. \]

Substituting the values of \( a^2 \) and \( b^2 \):

\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}. \]

### Step 3: Find the foci of the ellipse
The foci of the ellipse are located at \( (0, \pm c) \), where \( c \) is given by:

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

Thus, the foci are at the points \( (0, 4) \) and \( (0, -4) \).

### Step 4: Determine the distance from point \( P(x_1, y_1) \) to the foci
Let’s denote the foci as \( F_1(0, 4) \) and \( F_2(0, -4) \).

The distance \( d_1 \) from point \( P(x_1, y_1) \) to the focus \( F_1(0, 4) \) is given by:

\[ d_1 = \sqrt{(x_1 - 0)^2 + (y_1 - 4)^2} = \sqrt{x_1^2 + (y_1 - 4)^2}. \]

The distance \( d_2 \) from point \( P(x_1, y_1) \) to the focus \( F_2(0, -4) \) is:

\[ d_2 = \sqrt{(x_1 - 0)^2 + (y_1 + 4)^2} = \sqrt{x_1^2 + (y_1 + 4)^2}. \]

### Step 5: Calculate the distances
Thus, the distances from the point \( P(x_1, y_1) \) to the foci are:

\[ d_1 = \sqrt{x_1^2 + (y_1 - 4)^2} \]
and
\[ d_2 = \sqrt{x_1^2 + (y_1 + 4)^2}. \]