bisector theorem , cosine rule or other method? A fixed point $A(a,0)$, where $a>0$ and a straight line $l$ is given $x = -1$, $B$ is a moving point on the second quadrant and lies on the straight line $l$, the angle bisector of $\angle BOA$ intersects $AB$ at point $C$.
(a) Find the locus of point $C$ and state the range values of $x$ and $y$.
(b) Discuss the relationship between the values of $a$ and the type of the curve of the equation obtained in (a).

I have tried to use cosine rule to and bisector theorem to obtained the locus. but the equation seem too complicated. Any other good method please share with me !!

Let $B=(-1,t)$, $C=(x,y)$.
Then
$$\frac{y}{x-a}=\frac{t}{-1-a}$$
Angle AOB can be obtained to be
$$\cos AOB=\frac{-a}{a\sqrt{1+t^2}}=\frac{-1}{\sqrt{1+t^2}}$$
Using
$$\cos^2 AOC = \frac{1+\cos AOB}{2}$$
we have
$$\frac{y^2}{x^2}=\tan^2 AOC = \sec^2 AOC - 1=\frac{2}{1-\frac{1}{\sqrt{1+t^2}}}-1=\frac{2\sqrt{1+t^2}}{\sqrt{1+t^2}-1}-1=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}-1}=\frac{\left(\sqrt{1+t^2}+1\right)^2}{t^2}$$
$$\frac{y}{x}=\frac{\sqrt{1+t^2}+1}{t}$$
$$\left(t\frac{y}{x}-1\right)^2=1+t^2$$
$$\left(ty-x\right)^2=\left(1+t^2\right)x^2$$
$$\left(\frac{(1+a)y^2}{x-a}+x\right)^2=\left(1+\frac{(1+a)^2y^2}{(x-a)^2}\right)x^2$$
$$((1+a)y^2+x(x-a))^2=((x-a)^2+(1+a)^2y^2)x^2$$
$$(1+a)^2y^4+2(1+a)x(x-a)y^2=(1+a)^2x^2y^2$$
$$y^2+\frac{2}{1+a}x(x-a)-x^2=0$$
$$\left(\frac{2}{1+a}-1\right)x^2-\frac{2a}{1+a}x+y^2=0$$
$$(1-a)x^2-2ax+(1+a)y^2=0$$
If $0<a<1$, then it is an ellipse.
If $a=1$, then it is a parabola.
If $a>1$, then it is a hyperbola.
