Question

$ABC$ is an isosceles triangle with vertex at $A$ and $P$ is any point inside the triangle. If the rectangle contained by perpendicular from $P$ to sides $AB$ and $AC$ is equal to square of the perpendicular from $P$ to base $BC$, then prove that the locus of $P$ is a circle.

Answer 1

since the triangle is isosceles its median is also its right bisector. hence the vertices are marked with
$AB=AC=\sqrt{a^2+b^2}$
The equations of the sides by intercepts form are
$\frac{x}{a}+\frac{y}{b}-1=0,\frac{x}{-a}+\frac{y}{b}-1=0$
or $bx+ay-ab=0,-bx+ay-ab=0$
and base $BC$ is $y=0$. The constant term in both the equations of sides is kept of same sign as the point $P$ is inside.
If $P$ be $(x,y)$ then
$\frac{bx+ay-ab}{\sqrt{a^2+b^2}}.\frac{-bx+ay-ab}{\sqrt{a^2+b^2}}=y^2$
or ${(ay-ab)}^2-b^2{x}^2=(a^2+b^2)y^2$
or $x^2+y^2+\frac{2a^2}{b}y-a^2=0$
Above equation represents a circle whose centre is at $(0,-\frac{a^2}{b})$