Equation of a line joining a foci of the ellipse `x^2/25+y^2/9=1` to a foci of the ellipse `x^2/9+y^2/25=1` is ,

Options

A. x+y=4
B. x+y=5
C. x+y=3
D. 5x+3y=1

Answer

To find the equation of the line joining the foci of the ellipses given by the equations \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \) and \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \), we will follow these steps:

### Step 1: Identify the foci of the first ellipse
The first ellipse is given by:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
Here, \( a^2 = 25 \) and \( b^2 = 9 \). Since \( a > b \), the foci are located at:
\[ (\pm ae, 0) \]
where \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

Calculating \( e \):
\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus, the foci are:
\[ F_1 = (5 \cdot \frac{4}{5}, 0) = (4, 0) \quad \text{and} \quad F_2 = (-5 \cdot \frac{4}{5}, 0) = (-4, 0) \]

### Step 2: Identify the foci of the second ellipse
The second ellipse is given by:
\[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \]
Here, \( a^2 = 9 \) and \( b^2 = 25 \). Since \( b > a \), the foci are located at:
\[ (0, \pm be) \]
where \( e = \sqrt{1 - \frac{a^2}{b^2}} \).

Calculating \( e \):
\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus, the foci are:
\[ F_3 = (0, 5 \cdot \frac{4}{5}) = (0, 4) \quad \text{and} \quad F_4 = (0, -5 \cdot \frac{4}{5}) = (0, -4) \]

### Step 3: Find the equation of the line joining the foci
We will find the equation of the line joining the points \( F_1(4, 0) \) and \( F_3(0, 4) \).

Using the two-point form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \( (x_1, y_1) = (4, 0) \) and \( (x_2, y_2) = (0, 4) \):
\[ y - 0 = \frac{4 - 0}{0 - 4}(x - 4) \]
This simplifies to:
\[ y = -1(x - 4) \]
or
\[ y = -x + 4 \]
Rearranging gives:
\[ x + y = 4 \]

### Final Answer
The equation of the line joining the foci of the two ellipses is:
\[ \boxed{x + y = 4} \]