Locus of point M such that its projections onto two fixed lines are always at the same distance. A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is
$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$
What I have done: Suppose $P$ is any point on the locus, $A$ and $B$ be the feet of the perpendiculars, and $O$ is the origin. Then I want to show that
$$OP=\frac{AB}{\sin AOB}=\frac{c}{\sin \theta}$$
where $\theta$ is the angle between the two lines.
Note that the arc $PAO$ is a semi-circle, since $\angle PAO=90$. Similarly the arc $PBO$ is a semi-circle since $\angle PBO=90$. Also $\angle APB=\angle AOB$, as it is the angle of the same arc.
Now, $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. From here I can find $\sin \theta$ and $OP=\sqrt{x^2+y^2}$. Putting all these in,
$$OP=\frac{c}{\sin \theta}$$
will give me the equation of the locus. The problem is how to show that
$$OP=\frac{AB}{\sin AOB}.$$

To easily prove that the equation of locus whose equation is recognizable to be a circle of radius R we are given
$$4(x^2+y^2)(h^2-ab)=4 R^2 (h^2-ab)=c^2(4h^2+(a-b)^2) $$
i.e., we are given
$$  \frac{R^2}{c^2}=\frac{(4h^2+(a-b)^2)}{4(h^2-ab)} \tag 1 $$
Consider the special case when the moving point is directly on one of the pair of straight lines configured in this simplest way:

$$ \frac {c}{R} = \sin \alpha  \tag 2 $$
Angle included between straight line pair is known, given by:
$$ \tan \alpha =\frac{\sqrt{h^2-ab}}{(a+b)/2} \tag 3 $$
Elimination of $\alpha$ between equations (2) and (3) results in (1)...  done !
