(3) The graph $F_{1}$ of the function $y=x^{2}$ and the graph $F_{2}$ of the function after it is translated by the vector $\vec{a}=(m, 1)$ have perpendicular tangents at $x=1$, then the value of the real number $m$ is ( ).
(A) $-\frac{5}{4}$
(B) $-\frac{3}{4}$
(C) $\frac{3}{4}$
(D) $\frac{5}{4}$
(3) $\mathrm{D}$ Hint: Since $y^{\prime}(1)=2$, the function of $F_{2}$ is $y=(x-m)^{2}-1$, and its tangent line at $x=1$ has a slope of $k_{2}=2(1-m)$. From $2 \times 2(1-m)=$ -1, we know $m=\frac{5}{4}$.