Condition that the line $y=mx+c$ is normal to a conic I know that the condition for tangency can be derived by plugging $y=mx+c$ in the equation and setting $D=0$ since the line intersects at one point. But what condition can be used to derive the relation between c and m for the normal to a conic?
 A: Put $y=mx+c$ into the equation of the conic and solve the resulting equation for the value(s) of $x$. Note that you can get $0$,$1$ or $2$ values for $x$. Let those $x$ be $x_1$ and $x_2$, if they exist. Then, the slope of the conic at $(x_1,y_1)$ OR $(x_2,y_2)$ must be equal to $-\frac{1}{m}$.

As asked in the comments, let the conic be a standard hyperbola. Then, 
$$\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$$
$$(b^2-a^2m^2)x^2-2mca^2x-(a^2c^2+b^2a^2)=0$$
For it's roots to be real, $D>0$. Note that $D\geq 0$.
$$m^2c^2a^2+(a^2c^2+b^2a^2)(b^2-a^2c^2)\geq 0\\ \text{giving}$$
$$b\geq \sqrt{a^2m^2-c^2}$$
Note that it's just the condition that the given line will intersect the hyperbola at one or two points. You'll need to find the slopes at those points to assure that it actually is normal to it.
A: Let $f(x,y)$ be the polynomial so that $f(x,y)=0$ represents the given conic.
Let $(x_0,y_0)$ be a point where the given line intersects the conic. Then the tangent to the conic at that point has equation
$$
(x-x_0)\frac{\partial f}{\partial x}(x_0,y_0)+
(y-y_0)\frac{\partial f}{\partial y}(x_0,y_0)=0
$$
and the orthogonality relation becomes
$$
m\frac{\partial f}{\partial x}(x_0,y_0)-\frac{\partial f}{\partial y}(x_0,y_0)=0
$$
If the line is expressed as $ax+by+c=0$, then the orthogonality relation is
$$
a\frac{\partial f}{\partial x}(x_0,y_0)+b\frac{\partial f}{\partial y}(x_0,y_0)=0
$$
Just to show how this work in a particular case, let $f(x,y)=x^2-y$. The resolvent equation is $x^2-mx-c=0$; there are two distinct intersections (necessary condition for the line being normal) if and only if $m^2+4c>0$. Setting $r=\sqrt{m^2+4c}$, the intersection points are at $x=(m+r)/2$ and $x=(m-r)/2$, so we get the points
$$
\left(\frac{m+r}{2},\frac{m^2+2c+mr}{2}\right)
\qquad
\left(\frac{m-r}{2},\frac{m^2+2c-mr}{2}\right)
$$
The partial derivatives are
$$
\frac{\partial f}{\partial x}=2x \qquad \frac{\partial f}{\partial y}=-1
$$
Thus the orthogonality relation for the first point is
$$
m(m+r)+1=0
$$
that becomes
$$
r=-\frac{1}{m}-m
$$
and, squaring,
$$
m^2+4c=\frac{1}{m^2}+2+m^2
$$
that is,
$$
c=\frac{1+2m^2}{4m^2}
$$
The final condition is the same using the other point.
