(2) Given that point $P$ on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ is 4 units away from the point $(4,0)$, then the distance from point $P$ to the line $x=-\frac{25}{4}$ is ( ).
(A) 4
(B) 6
(C) $\frac{15}{2}$
(D) $\frac{5}{4}$
(2) C Hint: Since $a=5, b=3$, then $c=4$. Thus, the distance from $P$ to the other focus $(-4,0)$ is $2 \times 5-4=6$. Since the line $x=-\frac{25}{4}$ is the left directrix of the ellipse, the distance from $P$ to the line $x=-\frac{25}{4}$ is $d=\frac{6}{e}=7.5$.