locus of orthocentre of given triangle if $\tan B+\tan C = c$ 
Vertex $A$ of $\triangle ABC$ moves in such a way that $\tan B+\tan C = c(\bf{constant}),$
Then locus of orthocentre of $\triangle ABC$ is ( side $BC$ is fixed)

$\bf{My\; Try::}$ We can write $$\tan B+\tan C = c\Rightarrow \frac{\sin B}{\cos C}+\frac{\sin C}{\cos B} = c$$
So $$\frac{\sin (B+C)}{\cos B \cos C} = c\Rightarrow \sin (B+C)=c \cdot \cos B \cos C$$
Using $A+B+C = \pi\Rightarrow A+B = \pi-C$
So $$\sin (\pi-A) = c\cdot \cos B \cos C\Rightarrow \sin A = c\cdot \cos B\cdot \cos C$$
Now how can i solve it after that, Help required, Thanks
 A: 
Let $B=(-1,0)$, $\>C=(1,0)$ and $A=(u,v)$ with $-1<u<1$, $\>v>0$. The point $A$ then has to move such that
$${v\over 1+u}+{v\over 1-u}=c\ ,$$
which implies
$$2v=c(1-u^2)\ .\tag{1}$$
Let ${\pi\over2}-\beta=:\beta'>0$. The height $h_C$ then has the equation
$$h_C:\quad y=(1-x)\tan\beta'=(1-x){1+u\over v}\ .$$
This height intersects the height $h_A: \ x=u$ in the point $H=(u,z)$ with
$$z=(1-u){1+u\over v}={2\over c}\qquad(-1<u<1)\ ,$$
because of $(1)$. This means that when the point $A$ moves in a legal way then the orthocenter $H$ moves on a horizontal line at level ${2\over c}$.
A: Suppose $B(-a,0)$ and $C(a,0)$, suppose $A(x_A,y_A)$ (in the first quadrant for easyness), then, the relation $tan(b) + tan(c) =C$ translates into $\frac{y_A}{x_A+a}+\frac{y_A}{a-x_A}=C$. Performing calculations you will arrive to $y_A=\frac{-C}{2a}{x_{A}}^2+\frac{aC}{2}$. 
Now observe that $H$ is the intersection of a vertical line $(v):x=x_A$ with a perpendicular to $AB$ through $C$ of equation $(p):y=\frac{-1}{tan(b)}(x-a)=\frac{x_A+a}{y_A}(x-a)$. Solving the coordinates of $H$ gives $H(x_A, \frac{{x_A}^2-a^2}{y_A})$. Combining with the value $y_A$ above and eliminating variables in the coordinates of $H$ yields $$y= \frac{x^2-a^2}{\frac{-C}{2a}{x}^2+\frac{aC}{2}}$$ which implies the locus of $H$ is a curve that can be easily plotted since it is an elemental function of $x$.
