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

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 $x$.
