A question about Hartshorne Algebraic Geometry Chapter III Exercise 9.6 (I am not a native English speaker hence there may be some mistakes.)
Recently I was working on the problem III.9.6 in Hartshorne, Algebraic Geometry. It states:

Let $Y \subset \mathbb{P}^n$ be a nonsingular variety of dimension $\ge2$ over an algebraically closed field $k$. Suppose $\mathbb{P}^{n-1}$ is a hyperplane in $\mathbb{P}^{n}$ which does not contain Y, and such that the scheme $Y'=Y \cap \mathbb{P}^{n-1}$ is also nonsingular. Prove that Y is a complete intersection in $\mathbb{P}^{n}$ if and only if $Y'$ is a complete intersection in $\mathbb{P}^{n-1}$.

However, I was stuck on the sufficiency. I have tried to search for this problem but found no results. Hartshorne's book has a hint: use (9.12) applied to the affine cone of $Y$ and $Y'$. I have tried this hint, however (9.12) is an algebraic result concerning about Noetherian local ring. For affine cone of $Y'$, although its coordinate ring is normal since $Y'$ is complete intersection hence projectively normal, but it seems to have no connection with the local ring. Also, I really do not understand how this problem is related to section 9, Flat Families.
For the completeness, I post (9.12) here:

Let $A$ be a local noetherian domain, which is a localization of an algebra of finite type over a field $k$. Let $t \in A$, and assume:
(1). $tA$ has only one minimal associated prime ideal $\mathfrak{p}$,
(2). $t$ generates the maximal ideal of $A_{\mathfrak{p}}$,
(3). $A/\mathfrak{p}$ is normal.
Then $\mathfrak{p}=tA$ and $A$ is normal.

Finally, I really appreciate any people who can offer help to me. Thank you very much!
 A: We use the following corollary of Lemma of Hironaka:

Let $A$ be a noetherian domain (not necessarily local!), which is
  excellent (e.g. a localization of an algebra of finite type over a
  field $k$). 
Let $t \in A$, and assume:
(1). $tA$ has only one minimal associated prime ideal $\mathfrak{p}$,
(2). $t$ generates the maximal ideal of $A_{\mathfrak{p}}$,
(3). $A/\mathfrak{p}$ is normal.
Then $\mathfrak{p}=tA$.

Proof.
Let $\mathfrak{q} \subset A$ be any prime ideal.
Case 1. $t \notin \mathfrak{q}.$
$\mathfrak{p} A_{\mathfrak{q}} = A_{\mathfrak{q}} = tA_{\mathfrak{q}}.$
Case 2. $t \in \mathfrak{q}.$
In this case, the local ring $A_{\mathfrak{q}}$ satisfies the conditions of Lemma of Hironaka.
Therefore, $\mathfrak{p} A_{\mathfrak{q}} = t A_{\mathfrak{q}}.$
Summing up, at every prime $\mathfrak{q}$, $\mathfrak{p} A_{\mathfrak{q}} = t A_{\mathfrak{q}}.$
Hence $\mathfrak{p} = t A. \blacksquare$
Using this, we can conclude that $\operatorname{I}_{\mathbb{P}^n} (Y') = \operatorname{I}(Y) + (x_n).$
Since $\operatorname{I}_{\mathbb{P}^n} (Y') = \operatorname{I}_{\mathbb{P}^{n-1}} (Y') + (x_n)$, we can write 
$$\operatorname{I}_{\mathbb{P}^n} (Y') = (f_1, \cdots , f_r, x_n).$$
The rest is easy!
