Question

The sides AB and AC of a triangle ABC are given in position and the harmonic mean between the lengths AB and AC is also given; prove that the locus of the focus of the parabola touching the sides at B and C is a circle whose centre lies on the line bisecting the angle BAC.

Answer 1

Let $AB$ and $AC$ be the axes of length $a$ and $b$ respectively.

Let $k$ be the harmonic mean between $a$ and $b$

$\therefore \frac{1}{a}+\frac{1}{b}=\frac{2}{k}.......\left(i\right)$

Focus of the parabola is given by

$ax=by=x^2+{2xy\cos \omega +y}^2\Rightarrow a=\frac{x^2+{2xy\cos \omega +y}^2}{x},b=\frac{x^2+{2xy\cos \omega +y}^2}{y}$

Substituting $a$ and $b$ in $\left(i\right)$, we get

$\frac{x}{x^2+{2xy\cos \omega +y}^2}+\frac{y}{x^2+{2xy\cos \omega +y}^2}=\frac{2}{k}\frac{x+y}{x^2+{2xy\cos \omega +y}^2}=\frac{2}{k}\frac{k}{2}(x+y)=x^2+{2xy\cos \omega +y}^2{x}^2+{2xy\cos \omega +y}^2-\frac{kx}{2}-\frac{ky}{2}=0$

Clearly the equation represents circle with centre $(\frac{k}{4},\frac{k}{4})$

Now, $x$ and $y$ coordinate of the centre are same. So the centre lies on the line bisecting angle $BAG$.

Hence proved.