The foci of an ellipse are `S(3,1)` and `S'(11,5)`The normal at P is `x+2y-15=0` Then point P is

Options

A. a) 17,1
B. b) -17,-1
C. c) -17,1
D. d) 17,-1

Explanation

To solve the problem step by step, we will follow the given information and derive the required point \( P \).

### Step 1: Identify the foci of the ellipse
The foci of the ellipse are given as:
- \( S(3, 1) \)
- \( S'(11, 5) \)

### Step 2: Calculate the distance between the foci
The distance \( d \) between the foci \( S \) and \( S' \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(11 - 3)^2 + (5 - 1)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]

### Step 3: Relate the distance to the ellipse parameters
The distance between the foci is related to the semi-major axis \( a \) and the eccentricity \( e \) of the ellipse by the formula:
\[ d = 2ae \]
Thus, we have:
\[ 2ae = 4\sqrt{5} \implies ae = 2\sqrt{5} \]

### Step 4: Use the relationship between \( a \), \( b \), and \( e \)
We know that:
\[ c^2 = a^2 - b^2 \]
where \( c \) is the distance from the center to each focus. Since \( c = ae \), we can substitute:
\[ (ae)^2 = a^2 - b^2 \]
Substituting \( ae = 2\sqrt{5} \):
\[ (2\sqrt{5})^2 = a^2 - b^2 \implies 20 = a^2 - b^2 \]

### Step 5: Write the equation of the normal line
The normal line at point \( P(x_1, y_1) \) on the ellipse is given as:
\[ x + 2y - 15 = 0 \]

### Step 6: Set up the equations
From the normal line equation, we can express \( y \) in terms of \( x \):
\[ 2y = 15 - x \implies y = \frac{15 - x}{2} \]
Let \( y_1 = \frac{15 - x_1}{2} \).

### Step 7: Substitute into the ellipse normal equation
The general form of the normal equation for an ellipse is:
\[ \frac{a^2}{x_1} (x - x_1) + \frac{b^2}{y_1} (y - y_1) = 0 \]
We will compare coefficients from the normal line equation and the ellipse normal equation.

### Step 8: Compare coefficients
From the normal line \( x + 2y - 15 = 0 \), we can equate:
- Coefficient of \( x \): \( 1 \)
- Coefficient of \( y \): \( 2 \)

From the ellipse normal equation, we have:
- Coefficient of \( x \): \( \frac{a^2}{x_1} \)
- Coefficient of \( y \): \( \frac{b^2}{y_1} \)

Setting up the equations:
\[ \frac{a^2}{x_1} = 1 \quad \text{(1)} \]
\[ \frac{b^2}{y_1} = 2 \quad \text{(2)} \]

### Step 9: Solve for \( x_1 \) and \( y_1 \)
From equation (1):
\[ a^2 = x_1 \]
From equation (2):
\[ b^2 = 2y_1 \]

### Step 10: Substitute \( y_1 \) into the equations
Substituting \( y_1 = \frac{15 - x_1}{2} \) into \( b^2 = 2y_1 \):
\[ b^2 = 2 \left( \frac{15 - x_1}{2} \right) = 15 - x_1 \]

### Step 11: Substitute \( b^2 \) into the ellipse equation
Now substituting \( b^2 \) into the equation \( a^2 - b^2 = 20 \):
\[ x_1 - (15 - x_1) = 20 \]
\[ 2x_1 - 15 = 20 \implies 2x_1 = 35 \implies x_1 = 17.5 \]

### Step 12: Find \( y_1 \)
Substituting \( x_1 \) back into \( y_1 \):
\[ y_1 = \frac{15 - 17.5}{2} = \frac{-2.5}{2} = -1.25 \]

### Conclusion
The point \( P \) is \( (17.5, -1.25) \).