Find the equations to the normal lines of a curve where $x = 4$. 
Find the equations to the normal lines for the curve defined by
$$y^2+xy-x^2=5$$
In the points where $x=4$.

I'm not entirely sure how to even tackle this kind of problem, but let's see:
When $x=4$, we would have
$$y^2+4y-16=5$$
And we would have two solutions: $y=3$ and $y=-7$, so we need the equations for the normal lines that pass through $(4,3)$ and $(4,-7)$.
So I'm already confused. How come this curve has two images for $x = 4$?
I don't think I even understand what I am being asked.
 A: You are given an implicitly defined curve $F(x,y)=0$ and found points $(x_0,y_0)$ on this curve. The tangent at these points is defined by implicit differentiation
$$
∂_xF(x_0,y_0)·(x-x_0)+∂_yF(x_0,y_0)·(y-y_0)=0
$$
which also tells you the direction of the normal to the tangent and thus to the curve.
A: A curve doesn't have to be a function, implicitly defined curve is given by relation $$F(x,y)=0$$
Where $z=F$ is a function $F\colon \mathbb R^2\to\mathbb R$. We are slicing it by a surface $z=0$:

Thus, the curve may be multivalued in $Oxy$ plane. And it looks like the following:

The second thing is a derivative of this curve. When we have some function of one variable $f\colon\mathbb R\to\mathbb R$, its derivative is function of one variable too. But things get complicated when considering a curve $C$ — its derivative is given by
$$G\colon C\to\mathbb R$$
So every point of this curve has its image — the slope of a tangent at this point. In your case slopes at points intersecting with $x=4$ are given by $G(4,3)$ and $G(4,7)$.
A: 
See, there are two tangent lines for the circle at $x=0$.
