Finding lines of symmetry algebraically How would you determine the lines of symmetry of the curve $y=f(x)$ without sketching the curve itself?
I was working on the problem of finding the lines of symmetry of the curve given by:
$ x^{4}$+$ y^{4}$ = $u$ where u is a positive real number.
but I had to resort to doing a quick sketch. 
Is there an explicit condition for a line to be considered a line of symmetry of a curve?
Thanks.

By definition, G
  has line of symmetry $y=px+q$
  if for every point $(\alpha,\beta)\in G$
 , its reflection point is also in $G$
 .
By the Lemma below, its reflection point is $\left(\frac{2p(\beta-q)+(1-p^{2})\alpha}{p^{2}+1},\frac{(p^{2}-1)\beta+2p\alpha+2q}{p^{2}+1}\right)$
 .
Hence, a graph $G$
  has line of symmetry $y=px+q$
  if and only if:
$$(\alpha,\beta)\in G\implies \left(\frac{2p(\beta-q)+(1-p^{2})\alpha}{p^{2}+1},\frac{(p^{2}-1)\beta+2p\alpha+2q}{p^{2}+1}\right)$$
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Lemma. The reflection point of $(\alpha,\beta)$
  on the line $y=px+q$
  is $$\left(\frac{2p(\beta-q)+(1-p^{2})\alpha}{p^{2}+1},\frac{(p^{2}-1)\beta+2p\alpha+2q}{p^{2}+1}\right)$$
Proof. Let $y=-(1/p)x+r$
  be the line that is perpendicular to the line of symmetry and which runs through $(\alpha,\beta)$
 . Then $\beta=-(1/p)\alpha+r
  \implies
 r=\beta+\alpha/p$.
So the equation of that perpendicular line is $y=-(1/p)x+\beta+\alpha/p$
 .
The intersection point of this perpendicular line with the original line of symmetry is given by:
$$px+q= -(1/p)x+\beta+\alpha/p \\
(p+1/p)x= \beta+\alpha/p-q \\
x= \frac{p\beta+\alpha-pq}{p^{2}+1}=\frac{p(\beta-q)+\alpha}{p^{2}+1} \\
y= px+q \\
y= p\frac{p(\beta-q)+\alpha}{p^{2}+1}+q \\
y= \frac{p^{2}(\beta-q)+p\alpha+(p^{2}+1)q}{p^{2}+1} \\
y= \frac{p^{2}\beta+p\alpha+q}{p^{2}+1} $$
The reflection point is twice as far away from $(\alpha,\beta)$
  as is the above intersection point. So the reflection point is:
$$\left(2\frac{p(\beta-q)+\alpha}{p^{2}+1}-\alpha,2\frac{p^{2}\beta+p\alpha+q}{p^{2}+1}-\beta\right) \\
= \left(\frac{2p(\beta-q)+2\alpha-(p^{2}+1)\alpha}{p^{2}+1},\frac{2p^{2}\beta+2p\alpha+2q-(p^{2}+1)\beta}{p^{2}+1}\right) \\
= \left(\frac{2p(\beta-q)+(1-p^{2})\alpha}{p^{2}+1},\frac{(p^{2}-1)\beta+2p\alpha+2q}{p^{2}+1}\right) \blacksquare $$
