Prove quadratic equation ($y=ax^2+bx+c$) has only one line of symmetry We can show that the graph of the quadratic equation $y=ax^2+bx+c$ has the line of symmetry $x=-b/2a$. But how can we show that this is unique? (That is, why is no other line $dx+ey+f=0$ a line of symmetry?)
(I've been trying to show that no other line of symmetry can work but have simply been drowning in a sea of algebra. I imagine there must be some cleverer way to do it.)
 A: As Yves Daoust shows, there is  only one vertical line of symmetry. And a line that is not vertical intersects the parabola in $0,1,$ or $2$ points. If there is only one point of intersection, then the line is a tangent, which is not an axis of symmetry. Othwerise, the portion of the parabola on one side of such a line is either empty or of finite length. But the parabola has infinite length, so such a line can't be a line of symmetry.
A: Symmetry around $x=s$ is expressed by
$$ax^2+bx+c=a(2s-x)^2+b(2s-x)+c$$ for all $x$.
By identification of the powers of $x$, we get the compatibility condition
$$b+2as=0$$ or $$s=-\frac b{2a}.$$

Consider an arbitrary parabola and an arbitrary line. By rotation we can bring the line on the axis $y$. Then the equation of the rotated parabola
$$ax^2+2bxy+cy^2+2dx+2ey+f=0$$
 must be invariant to a change of the sign of $x$ and it must reduce to
$$ax^2+cy^2+2ey+f=0.$$
For this equation to represent a parabola, we must have $c=0$. Then,
$$ax^2+2ey+f=0$$ is a parabola for which the axis of symmetry is $y$. Hence any line which is an axis of symmetry is the standard axis.

Note that in the case $ac\ne0$ (centered conic), by translation we can further reduce to
$$ax^2+cy^2+f=0$$ and the line is one of the known axis of symmetry. Hence there are exactly two of them.
A: Any line of symmetry of a parabola must intersect all chords perpendicular to it at their midpoints. A property of parabolas is the the locus of the midpoints of a family of parallel chords is a line parallel to the parabola’s axis. The tangent at the intersection of this line and the parabola is also parallel to the chords. The only point on the parabola at which the tangent is perpendicular to its axis is its vertex, so there’s only one line of symmetry—the parabola’s axis $x=-b/2a$. Depending on how you define the parabola’s axis, this argument could end up being circular.  
Alternatively, the proof can be bashed out algebraically. Take an arbitrary line $lx+my+n=0$. W.l.o.g. we can assume that $l^2+m^2=1$ in order to simplify the calculations. Reflection in this line then amounts to making the substitutions $$x \to (1-2l^2)x-2lmy-2ln \\ y \to -2lmx + (1-2m^2)y-2mn.$$ We want to end up with a multiple of the original equation. In particular, the coefficient of $y^2$ must vanish. This comes from the $ax^2$ term in the original equation, from which, after squaring, we obtain the condition $$4al^2m^2=0.$$ Since $a\ne0$, we must have either $l=0$ or $m=0$, but not both. That is, the only possible lines of symmetry are horizontal or vertical.  
If $l=0$, then the reflection reduces to substituting $(1-2m^2)y-2mn$ for $y$, but then we must have $y=(1-2m^2)y-2mn$, which means that either $m=0$, which is not allowed, or $n=-my$, which is not a constant. Therefore $l=0$ yields no lines of symmetry.  
On the other hand, if $m=0$, reflection reduces to substituting $(1-2l^2)x-2ln$ for $x$. Expanding, equating coefficients and solving the resulting system of equations yields three solutions: $l=0$, which we reject; $l=1$, $n=b/2a$, i.e., the line $x=-b/2a$; and $l=-1$, $n=-b/2a$, which gives the same line.
