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A given problem is

If the graph of the parabola $y=x^2$ is translated to a position such that its $x$ intercepts are $−d$ and $e$, and its $y$ intercept is $−f$, (where $d,e,f>0$), show that $de=f$.

The example solution is

Since the $x$ intercepts are $-d$ and $e$ the parabola must be of the form $y=a(x+d)(x-e)$. Also since we only translated $y=x^2$ it follows that $a=1$. Now setting $x=0$ we have $-f=-de$ and the results [sic] follows.

Ok, I checked the solution but still do not understand. Can some please explain in the most succinct way? Why do we automatically assume that $a = 1$ when it could have been $2,3,4,...$?

This comes from the fact that we are only translating the parabola $y = x^2$. This means that the only operations to get to the new parabola are horizontal or vertical shifts (these would be replacing $x$ with $x+h$ or $y$ with $y+k$, respectively). Neither of these changes the coefficient on $x^2$.

Intuitively, changing $a$ will affect the curvature of the parabola — how quickly it gets steep. But translation means that we don’t want to alter this at all, just “slide” the parabola around rigidly.

We assume that $a=1$ because the coefficient of the $x^2$ term in $y=x^2$ is $1$.

Let me elaborate.

The solution makes the most general assumption about the parabola: we assume that $y$ denotes a parabola with $x$-intercepts $-d$ and $e$. Then we know that

$$y = a(x - (-d))(x - e)$$

We also know that we are considering $y = x^2$. Translation of $y$ does not change the coefficient of the $x^2$ term. (Play around with a graphing calculator. You'll notice that this term's coefficient only affects how "skinny" or "wide" the parabola is. That is, translating the graph affects the linear/constant terms, but not the $x^2$ term.)

Thus,

$$x^2 + ?x + ?= a(x - (-d))(x - e)$$

We multiply out the right side, and thus,

$$1x^2 + ?x + ? = ax^2 + a(d-e)x - ade$$

Notice why $a$ must be $1$? It's because the coefficients of the quadratic terms must be equal, i.e. $a=1$. (Similarly, we can also see that the linear terms and constant terms must be equal but because of the translation we don't know these, which is why I denoted them with $?$'s.)

Our translation of the graph didn't affect the term.

To elaborate further, or offer an alternate view, there is an alternate equation for a parabola,

$$y = a(x-h)^2 + k = ax^2 - 2hax + ah^2 + k$$

where $(h,k)$ is the vertex. You could make the argument then that translating the graph only affects the location of the vertex, since it's not been stretched or widened. Notice again: once multiplied out, we see that the quadratic term of $y$ has neither a $h$ or $y$ factor, i.e. the quadratic term is completely indifferent to the vertex of the parabola! Thus, translation does not at all affect the $ax^2$ term!

Thus, since we are only translating the parabola $y =x^2$, we have no problem and letting $a=1$!

the three expressions of a parabola are:
1) y=a(x-h)^2+k;
2) y=a(x+d)(x-e);
3) y=ax^2+bx-f;
But since no Dilation is mentioned just Translation so a=1

Equations 2) and 3) leads to de=f.