If $F_1$ and $F_2$ are the feet of the perpendiculars from the foci $S_1$ and $S_2$ of an ellipse $\frac{x^2}{5} + \frac{y^2}{3} = 1$ on the tangent at any point $P$ on the ellipse,then $(S_1 F_1) (S_2 F_2)$ is equal to
- A$2$
- B$3$
- C$4$
- D$5$
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