If `F_1 and F_2` be the feet of the perpendiculars from the foci `S_1 and S_2` of an ellipse `(x^2)/5 + (y^2)/(3) = 1` on the tangent at any point P on the ellipse then `(S_1F_1) (S_2F_2)` is equal

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the product of the distances from the foci of the ellipse to the feet of the perpendiculars dropped from the foci onto the tangent at any point on the ellipse.

### Step-by-Step Solution

1. **Identify the Ellipse Parameters**:
The given ellipse is \(\frac{x^2}{5} + \frac{y^2}{3} = 1\).
Here, \(a^2 = 5\) and \(b^2 = 3\).
Thus, \(a = \sqrt{5}\) and \(b = \sqrt{3}\).

2. **Calculate the Foci**:
The foci of the ellipse are located at \(S_1 = (-c, 0)\) and \(S_2 = (c, 0)\), where \(c = \sqrt{a^2 - b^2}\).
\[ c = \sqrt{5 - 3} = \sqrt{2} \]
Therefore, the coordinates of the foci are \(S_1 = (-\sqrt{2}, 0)\) and \(S_2 = (\sqrt{2}, 0)\).

3. **Use the Formula for the Product of Perpendiculars**:
The product of the distances from the foci to the feet of the perpendiculars dropped from the foci onto the tangent at any point \(P\) on the ellipse is given by:
\[ S_1F_1 \cdot S_2F_2 = b^2 \]
Here, \(b^2 = 3\).

4. **Conclusion**:
Thus, we find that:
\[ S_1F_1 \cdot S_2F_2 = 3 \]

### Final Answer:
\((S_1F_1)(S_2F_2) = 3\)

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