<s>[INST] 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
$$OP=\frac{AB}{\sin AOB}.$$
 [/INST] Here's a non-trigonometric take on this problem.
The conic $ax^2+2hxy+by^2=0$ gives the two solutions $y=m_\pm x$ where $bm_\pm=-h\pm\sqrt{h^2-ab}$. Given $P(x_0,y_0)$, the perpendicular foot $A(x,m_+x)$ minimises $AP^2=(x-x_0)^2+(m_+x-y_0)^2$. This is minimised at $(m_+^2+1)x^*=x_0+m_+y_0$ so $$A\left(\frac{x_0+m_+y_0}{m_+^2+1},\frac{m_+(x_0+m_+y_0)}{m_+^2+1}\right)\implies B\left(\frac{x_0+m_-y_0}{m_-^2+1},\frac{m_-(x_0+m_-y_0)}{m_-^2+1}\right).$$ without loss of generality. As $AB=c$ we have $c^2=(\alpha^2+\beta^2)(x_0^2+y_0^2)$ where $\alpha=\dfrac1{m_+^2+1}-\dfrac1{m_-^2+1}$ and $\beta=\dfrac{m_+}{m_+^2+1}-\dfrac{m_-}{m_-^2+1}$. Now \begin{align}\alpha^2+\beta^2&=\frac1{m_+^2+1}+\frac1{m_-^2+1}-\frac{2(m_+m_-+1)}{(m_+^2+1)(m_-^2+1)}\\&=\frac{(m_+-m_-)^2}{(m_+m_-)^2+m_+^2+m_-^2+1}\\&=\frac{4(h^2-ab)/b^2}{a^2/b^2+2(2h^2-ab)/b^2+1}\\&=\frac{4(h^2-ab)}{4h^2+(a-b)^2}\end{align} using Vieta so the locus is a circle centred at the origin of radius $\dfrac c2\sqrt{\dfrac{4h^2+(a-b)^2}{h^2-ab}}$.
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