# Is the dual of the normal bundle of the exceptional divisor ample

Let $X$ be a non-singular projective variety and $Y \subset X$ a non-singular subvariety. Denote by $\tilde{X}$ the blow-up of $X$ along $Y$ and $E$ the associated normal bundle. Denote by $N_{E|\tilde{X}}$ the normal bundle of $E$ in $\tilde{X}$. We know that $N_{E|\tilde{X}} \cong \mathcal{O}_E(-1)$. Does this mean that the dual of the normal bundle $N_{E|\tilde{X}}$ is ample?

The bundle $O_E(1)$ is always relatively ample over $Y$. It is ample if and only if the conormal bundle $N^\vee_{Y/X}$ is ample (this is by definition). For instance, if $Y = \mathbb{P}^1$, this holds if and only if $N^\vee \cong \oplus \mathcal{O}(a_i)$ with $a_i > 0$ for all $i$.
This need not be true. The projective bundle version $\mathcal O_{\mathbb P(\mathcal E)}(1)$ does not have the same automatic ampleness properties as the ample generator on projective space.
For instance, you can think of $\mathbb P^1 \times \mathbb P^1$ as the projectivization of $\mathcal E := \mathcal O_{\mathbb P^1} \oplus \mathcal O_{\mathbb P^1}$ and $\mathcal O_{\mathbb P(\mathcal E)}(1)$ will be $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,0)$ (meaning it has degree $1$ on the fibers but is trivial on sections of the projective bundle). On the other hand, you can obtain the same surface by projectivizing any twist $\mathcal E(k)$ and the resulting $\mathcal O(1)$ will be (if I recall correctly) $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,k)$ which is ample for $k > 0$.
To relate this back to your situation: although the isomorphism class of the exceptional divisor only depends on the class of $N_{Y|X}$ up to twisting by any line bundle on $Y$, the ampleness of $\mathcal O_{E}(1)$ will depend on how positive the conormal bundle to $Y$ was in the first place.