If diagonal points of a square are sliding on coordinate axes, locus of other two points

Question

The full question is shown above. We have to try and write coordinates of A and C in terms of a single variable which we can later eliminate, but I'm unable to accomplish this.

I tried using parametric equation of lines from the centre of square because we know A, C are $\sqrt{2}$ units away from midpoint. Let $(a,b)$ be the midpoint of the square at any time and line $AC$ make angle $\theta$ with positive x axis. We can get coordinates of AC from parametric form as shown $$\frac{x-a}{\cos\theta}=\frac{y-b}{\sin\theta}=\pm \sqrt{2}$$ Still need a relation between $a,b$. If we write equations of line BD using given information and find mid point of x and y intercept (which is centre of square $\equiv (a,b)$), we further get the relation $a=b\tan\theta$.

Clearly, we still have two variables $b,\theta$, thus, its not possible to find locus. Please provide an extra relation or an easier method to figure out the locus.

Answer 1

It is not difficult to prove (see figure below) that $\angle BOA=45°$. And one can prove, analogously, that $\angle BOC=45°$.

Hence $A$ and $C$ lie on the lines bisecting the quadrants, $y=x$ and $y=-x$.

Answer 2

You've made a good start, but since $a$ and $b$ are used in the equation $ax^2 + hxy + by^2 = 0$, and there's no apparent indication that $(a,b)$ is meant to be the midpoint of the square, you should probably use different variables for that. Nonetheless, I'm using $a$ and $b$ below as you did, for consistency with your work. One other point is that $\lvert AC\rvert = \sqrt{2}$, so $A$ and $C$ are actually $\frac{1}{\sqrt{2}}$ away from the center point, not $\sqrt{2}$.

For the extra relation, let the midpoint of $AC$ be $M$, then draw a horizontal line to the left, with it meeting the $y$-axis at the point $E$. Also, have the origin be $O$. Then $\triangle DEM \sim \triangle DOB$, with the similarity ratio being $2$, so $D$ has co-ordinates $(0,2b)$ and $B$ has co-ordinates of $(2a,0)$. Since $\lvert DB\rvert = \sqrt{2}$ then, using the Pythagorean theorem with $\triangle DOB$, we get

$$4a^2 + 4b^2 = 2 \;\;\to\;\; a^2 + b^2 = \frac{1}{2}$$