The sum of the focal distances of any point on the conic `(x^(2))/(25)+(y^(2))/(16)` =1 is

Options

A. 10
B. 9
C. 41
D. 18

Answer

To solve the problem, we need to determine the sum of the focal distances of any point on the given conic section, which is an ellipse defined by the equation:

\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 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, we can identify:
- \( a^2 = 25 \) which gives \( a = \sqrt{25} = 5 \)
- \( b^2 = 16 \) which gives \( b = \sqrt{16} = 4 \)

### Step 2: Determine the length of the major axis
The length of the major axis of an ellipse is given by \( 2a \).

Calculating this:
\[ \text{Length of major axis} = 2a = 2 \times 5 = 10 \]

### Step 3: Understand the sum of the focal distances
For any point \( P \) on the ellipse, the sum of the distances from that point to the two foci \( S \) and \( S' \) (denoted as \( S_P + S'_P \)) is equal to the length of the major axis.

### Step 4: Conclusion
Thus, the sum of the focal distances of any point on the ellipse is:

\[ S_P + S'_P = 10 \]

### Final Answer
The sum of the focal distances of any point on the conic \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) is **10**.

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