if S and S are two foci of an ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` (`altb`) and `P(x_(1) , y_(1))` a point on it then SP+ S'P is equal to

Options

A. 2a
B. 2b
C. `a+ex_(1)`
D. `b+ey_(1)`

Answer

To solve the problem, we need to find the sum of the distances from a point \( P(x_1, y_1) \) on the ellipse to the two foci \( S \) and \( S' \). The equation of the ellipse is given as:

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

where \( b > a \).

### Step-by-Step Solution:

1. **Identify the Foci of the Ellipse**:
For an ellipse of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \( b > a \), the foci are located at:
\[ S(0, c) \quad \text{and} \quad S'(0, -c) \]
where \( c = \sqrt{b^2 - a^2} \).

2. **Understand the Property of Ellipses**:
A fundamental property of ellipses states that for any point \( P \) on the ellipse, the sum of the distances from that point to the two foci is constant. This constant is equal to \( 2b \) when the major axis is along the y-axis (as in our case).

3. **Apply the Property**:
Since \( P(x_1, y_1) \) lies on the ellipse, we can apply the property:
\[ SP + S'P = 2b \]

4. **Conclusion**:
Therefore, the sum of the distances from point \( P \) to the foci \( S \) and \( S' \) is:
\[ SP + S'P = 2b \]

### Final Answer:
\[ SP + S'P = 2b \]