3. Draw the tangent lines to the parabola $y=x^{2}+a x+4 a^{2}$ from the origin $O$. When $a$ varies, the two points of tangency lie on the parabolas ( ).
(A) $y=\frac{1}{2} x^{2}, y=\frac{3}{2} x^{2}$
(B) $y=\frac{3}{2} x^{2}, y=\frac{5}{2} x^{2}$
(C) $y=x^{2}, y=3 x^{2}$
(D) $y=3 x^{2}, y=5 x^{2}$
3.B.

Let the required tangent line equation be $y=k x$. Substituting it into the parabola $y=x^{2}+a x+4 a^{2}$, we get
$$
x^{2}+(a-k) x+4 a^{2}=0 .
$$

By $\Delta=(a-k)^{2}-16 a^{2}=0$, which means
$$
(k+3 a)(k-5 a)=0,
$$

we solve to get $k=-3 a, k=5 a$.
If $k=-3 a$, then the x-coordinate of the tangent point is $x=-2 a$. Substituting it into the tangent line equation $y=-3 a x$, we get the y-coordinate of the tangent point as $y=6 a^{2}$, thus, one tangent point lies on the parabola $y=\frac{3}{2} x^{2}$.
Similarly, the other tangent point lies on the parabola $y=\frac{5}{2} x^{2}$.