Suppose $X$ is a regular surface, i.e. $H^1(X)=0$, e.g. a toric surface. Then $Y$ is a smooth rational curve. So $X$ has a very ample smooth rational curve. This immediately requires $X$ to be rational by taking a pencil of smooth rational curves. Does every rational surface have such a curve? It's ok for projective space (lines) and Hirzebruch surfaces (high degree sections).
Not for $\mathbb P^2$ blown up at $k$ points for $k \geq 2$. Let $D = n H - \sum_{i=1}^k m_i E_i$. Then $D \cdot D = n^2 - \sum_{i=1}^k m_i^2$ and $K = -3 H + \sum_{i=1}^k E_i$ so $D \cdot K = -3n + \sum_{i=1}^k m_i$
By adjunction we have
$$-2= D \cdot D + D \cdot K = n^2 - \sum_{i=1}^k m_i^2 -3n + \sum_{i=1}^k m_i$$
$$ = \left(n- \sum_{i=1}^k m_i\right) \left(n + \sum_{i=1}^k m_i\right) +\sum_{1 \leq i< j \leq k} m_i m_j -3n + \sum_{i=1}^k m_i$$
$$ = \left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j -2$$
Thus
$$\left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j=0$$
But in fact it is positive:
$\left(n- \sum_{i=1}^k m_i-1\right)>0$ by ampleness because it is the intersection number with the line through the points.
$m_i\geq 1$ by ampleness because it's the intersection number with $E_i$, so $\left(n + \sum_{i=1}^k m_i-2\right) >0$ and $m_im_j>0$
This is a contradiction so this is impossible. When $k=2$ this variety is toric, showing that being toric is not a sufficient condition.