This problem is found in S.L. Loney's book on The Elements of Coordinate Geometry. The problem is pronounced in full as follows:
The base BC (=2a) of a triangle ABC is fixed; the axes being BC and a perpendicular to it through its middle point, find the locus of the vertex A, when the difference of the base angles is given (=a)
The way I attempted at it was: I realised that once the tangent of one of the angles was $=$$y\over x + a$ then the other would be equal to $=$$y\over x — a$. Hence let the tangent of one of the base angles, B (say), be $y\over x + a$; and let the tangent of C then be $y\over x — a$. And it is given that $B — C = a$. Taking the tangent of both sides of the last equation, we get: $\tan (B — C) =$$\tan B — \tan C \over 1 + \tan B \tan C$. Since we know the values of all these expressions we can substitute to get:
$\tan a =$$—2ay \over x^2 + y^2 — a^2$
Algebraically rearranging this equation we get
$x^2\tan a + y^2\tan a + 2ay — a^2\tan a = 0$
or much more concisely:
$\tan a(x^2 + y^2 — a^2) + 2ay = 0$
I am still knew to the grand picture of Coordinate Geometry, hence I seek affirmation from the more profesional of the answer I have provided to satisfy the problem. If I have made any fallacy in reasoning, reader, please say so and show me how its properly done. Thank you in advance.
We assume that $B$ and $C$ are on the $x$-axis (as opposed to the $y$-axis), and that $B$ is on the right of $C$.
We further assume that point $A$ is in the first quadrant. So $\angle{B}-\angle{C}=a$.
For convenience, I will write $B$ instead of $\angle{B}$, and $C$ instead of $\angle{C}$.
$$\tan{B}=\frac{y}{a-x}$$
$$\tan{C}=\frac{y}{a+x}$$
$$\tan{(B-C)}=\tan{a}$$
$$\implies \frac{\frac{y}{a-x}-\frac{y}{a+x}}{1+(\frac{y}{a-x})(\frac{y}{a+x})}=\tan{a}$$
$$\implies x^2-y^2+\frac{2}{\tan{a}}xy-a^2=0 \text{ with } x>0 \text{ and } y>0$$
Alternatively, we could have assumed that point $A$ is in any of the other three quadrants. Accordingly, we reflect the curve in the $x$-axis and $y$-axis. That is, we add absolute value signs around every $x$ and $y$, but note that they don't make a difference in $x^2$ and $y^2$.
So our final solution is
$$x^2-y^2+\frac{2}{\tan{a}}|xy|-a^2=0$$