Isosceles Triangle With Height Limiting To Zero, part 2 The figure shows an isosceles triangle ABC
with ∠B=∠C . The bisector of angle B intersects the side AC at the point P. Suppose that the base BC remains fixed but the altitude |AM| of the triangle approaches 0, so A approaches the midpoint M of BC.
Question: What is the shape of the trace of P as |AM| goes to 0?  Someone from https://answers.yahoo.com/question/index?qid=20130228222650AALgMhb says its a circle and someone else says its an ellipse.  My function describes neither.  Does anyone else get a circle or ellipse?

This function is $y=\sqrt(3x^2+2x-1)$.In order to deduce this equation write both the $x$ and $y$ position of the point p by writing the equations of lines in the coordinate with origin of M and intersect them.You should write the equations of line by considering $\tan(B/2)$ and $-\tan(B)$ as their slopes. the final values of x and y will be according $\tan(B)$ and $\tan(B/2)$. now if you delete the parametric variable B between x and y the equation will be achieved. Use $tan(2x)=2\tan(x)/1-\tan^2(x)$  
