Intersecting two pencils of plane curves In $\Bbb{P}^2$, let $D_1,D_2$ be two curves of degree $d_1,d_2$ respectively. Choose two pencils $|D_1(t)|\subset|D_1|$ and $|D_2(t)|\subset|D_2|$ (free of fixed components) parametrized by the same parameter $t\in\Bbb{P}^1$ such that


*

*The general member of the two pencils $|D_1(t)|$, $|D_2(t)|$ is irreducible.

*The two pencils have no basepoints in common.

*The parameter is chosen such that the curves belonging to both pencils (hence passing through the basepoints of both) are given by distinct values of $t$ in the two pencils.


Now consider the rational curve (Edit this is not rational: see DCV's comment below)
$$C:=\bigcup_t D_1(t)\cap D_2(t)$$
Notice $\deg(C)=d_1+d_2$. Simple questions arise:

Is $C$ irreducible/smooth? If not always, under which conditions?

Edit: as DCV suggests irreducibility does not always hold. What if we add the condition that the two pencils be general enough, i.e. each has the maximal number $(d_i)^2$ of distinct basepoints?
 A: We cannot show it, because it is not true. :)
Take $D_1(t): z -ty = 0$ and $D_2(t): x^2 + xz - 2z^2 -t(x^2 - xy - 2yz) = 0$, of degree $d_1 = 1$ and $d_2 = 2$.
The pencil $D_1(t)$ has one basepoint: $P=[1:0:0]$, while $D_2$ has two basepoints: $Q_1 = [0:1:0]$ and $Q_2 = [3:1:3]$. 
All conditions that you require are satisfied, but the curve $C$ is the union of $x = 0$ and $xy - xz + 2yz = 0$.
To check the third condition, note that $D_2$ passes through $P$ for $t = 1$, but $D_1(1):z-y = 0$ contains neither $Q_1$ nor $Q_2$.
A: The curve $C$ is neither necessarily irreducible, nor (a fortiori) smooth.
Suppose $D_1(t): f_1 - tf_2 = 0$ and $D_2(t): g_1 - tg_2 = 0$, where $f_1,f_2$ have degree $d_1$ and $g_1,g_2$ have degre $d_2$. Then, $C$ is given by $f_1g_2 - f_2g_1 = 0$.
Suppose $C$ contains a singular point $P$. Up to change of coordinates, we can suppose that $P$ is contained in $D_1(0)\cap D_2(0)$, which means $f_1(P) = 0$ and $g_1(P)=0$. The point $P$ is a singular point of $C$ if and only if
$$ g_2 \partial_x f_1 = f_2 \partial_x g_1,$$  $$g_2 \partial_y f_1 = f_2 \partial_y g_1,$$ $$g_2 \partial_z f_1 = f_2 \partial_z g_1.$$
For example, if $f_2(P) = 0$ (i.e., $P$ is a basepoint of $D_1(t)$), then $P$ is singular iff either $P$ is a basepoint of $D_2(t)$, too (excluded by your hypothesis), or $P$ is a singular point of $f_1 = 0$ (not excluded). Similarly if $g_2(P) = 0$.
If $f_2(P),g_2(P) \neq 0$ (i.e., $P$ is not a basepoint of either pencils), then $P$ is singular if $f_1$ and $g_1$ intersect with multiplicity at least 2 in $P$ (this is a necessary condition, but not sufficient!).
Anyway, it is easy to find an example with $d_i^2$ distinct basepoints for each pencil and $C$ reducible. For example, if $d_1 = 1$ and $d_2 =2$, then $\deg(C) = 3$ and if $C$ has two singular points then it is reducible.
Let $f_1 = y$, $f_2 = z$, $g_1 = 2x^2 - 2xy + 2y^2 - 3yz$, $g_2 = 2x^2 - 3xz + 2yz - 3z^2$. Then $D_1(t)$ has one basepoint $P = [1:0:0]$ and $D_2(t)$ has 4 basepoints $Q_1 = [0:3:2]$, $Q_2 = [3:3:2]$, $Q_3 = [1:-1:2]$, $Q_4 = [1:2:1]$. It satisfies all the conditions that you require, but $C$ is reducible and given by the union of $x = 0$ and $2xy - 2xz - yz = 0$.
