If y = x + c is a normal to the ellipse `x^2/9+y^2/4=1` , then `c^2` is equal to

Options

A. `13/25`
B. `25/13`
C. `25/9`
D. `13/4`

Answer

To solve the problem, we need to find the value of \( c^2 \) for the normal line \( y = x + c \) to the ellipse given by the equation \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \).

### Step-by-Step Solution:

1. **Identify the parameters of the ellipse**:
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
From the given equation \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \), we can identify:
\[ a^2 = 9 \quad \text{and} \quad b^2 = 4 \]

2. **Determine the slope of the normal line**:
The normal line is given by \( y = x + c \). Here, the slope \( m \) of the line is:
\[ m = 1 \]

3. **Use the formula for \( c \)**:
The formula for \( c \) when a line \( y = mx + c \) is normal to the ellipse is:
\[ c = \pm \frac{a^2 - b^2}{\sqrt{a^2 + b^2} \cdot m} \]
Substituting the values of \( a^2 \), \( b^2 \), and \( m \):
\[ c = \pm \frac{9 - 4}{\sqrt{9 + 4} \cdot 1} \]
Simplifying this gives:
\[ c = \pm \frac{5}{\sqrt{13}} \]

4. **Calculate \( c^2 \)**:
To find \( c^2 \):
\[ c^2 = \left(\pm \frac{5}{\sqrt{13}}\right)^2 = \frac{25}{13} \]

### Final Answer:
Thus, the value of \( c^2 \) is:
\[ \boxed{\frac{25}{13}} \]