Changing the direction of the homomorphism in the definition of pushouts In the diagram below $P$ is a pushout of the data if a unique $u:P\rightarrow Q$ exists for every solution $(Q,j_1,j_2)$ (by the definition of bushouts).
My question is : Suppose that the definition was: $P$ is a pushout of the data  $(Z,X,Y)$ if there exists a unique $u:Q\rightarrow P$ s.t $uj_1=i_1$ and $uj_2=i_2$. Then why it is not true or at least not  interesting definition?

 A: Added.
On universal properties in general. 
To say that a certain collection of data $U$ (objects and arrows in $\mathcal C$, say) is universal with respect to a property $\mathcal P$ means that whenever another collection of data $V$ (of the same nature as $U$) has $\mathcal P$, it must factor uniquely through $U$. 
NB: The fashion in which this happens, and which makes precise the sentence "$V$ factors through $U$", depends on $\mathcal P$.
Example (to clarify this dependence): the kernel. Let $f:A\to B$ be a morphism in $\mathcal C$. Now look at objects $X\in\textrm{Ob}\,\mathcal C$ satisfying the property $\mathcal P$: having a morphism $i:X\to A$ such that $f\circ i=0$.
The kernel of $f$ (when it exists) is by definition the universal couple $(K,j)$ with respect to $\mathcal P$. This means:

if another couple $(X,i)$ has $\mathcal P$, then there is a unique
  morphism $u:X\to K$ such that $i=j\circ u$.

Moral: what really matters is that we require a factorization of $i$ through $j$, not conversely! (and, by chance, now the morphism $u$ goes in the direction of $K$: but this is because of $\mathcal P$! it is $\mathcal P$ that asks for this to happen.)

The pushout of the diagram $$X\overset{f}{\longleftarrow}Z\overset{g}{\longrightarrow}Y$$ is the triple $(P;i_1,i_2)$ universal with respect to making the square you've drawn commutative. "Universal" means that if an extraneous object $(Q;j_1,j_2)$ attempts to imitate $(P;i_1,i_2)$, then it must factor (uniquely) through the universal object: in this context, it means that $j_1,j_2$ are obtained by "passing through" the universal object, i.e. by means of a (unique) map $u:P\to Q$. This is the only possible direction because otherwise it would be the universal object which factors through a non universal one.
So, the answer is: by definition of universal property.
[As an aside: this is the kind of argument that one uses to show that if two objects satisfy the same universal property, then they are isomorphic via a unique isomorphism.]

Also added.
The modified definition does not give anything new (i.e. different from the original square $i_2g=i_1f$). So to give that definition is the same as to give that commutative square. (Therefore the modified definition does not define a universal construction.)
Let us use the definition as in the question: we factor the universal arrows $i_1,i_2$ through "external" ones.
So let us take a triple $(Q;j_1,j_2)$ such that $j_1f=j_2g$. Then we find a unique $u:Q\to P$ such that $i_1=uj_1$ and $i_2=uj_2$. But notice that by considering $(Q;j_1,j_2)$ we gained nothing: the new commutative square, plus the "universal property" do not say anything new:
$$i_1f=(uj_1)f=u(j_1f)=u(j_2g)=(uj_2)g=i_2g,$$
which was the content of the original square. 
A: The modified definition you propose makes sense, it's just not a very interesting concept. Say you had a category with a final object, that is, an object $Z$ such that for any other object $X$, there is a unique morphism $X \to Z$. Then, with the modified definition you propose, every single pushout would just be $Z$.
Added: in a comment on atricolf's answer you mentioned your teacher told you to think of the case of the trivial group to see the problem. That's exactly what I was getting at above: in the category of groups, the "flipped" definition of pushout would make all pushouts equal to the the trivial group, which is not very useful.
