Finding a such that a parametric equation has solution

Question

I've been trying to solve this parametric equation: Find all $a$ for which there exist $x$ and $y$ such that $a(3x+2y) = y + \sqrt{2a^2(4x^2+y^2)} + \sqrt{2a^2x^2+2(a-1)^2y^2}$. I noticed x and y can scale, but trying to assume e.g. $x + y = 1$ has led me nowhere. I can't see any other substitution that would help me. Basic inequalities (AM-GM, Jensen, Cauchy-Schwarz) have also led me nowhere. After all this I have taken a look at the graph of the function on desmos and it looks like there are no such $a$. But how does one prove it?

Answer 1

Hint.

Making $y = \lambda x$ and substituting into

$$ a (3 x+2 y)-\sqrt{2} \sqrt{a^2 \left(4 x^2+y^2\right)}-\sqrt{2 a^2 x^2+2 (a-1)^2 y^2}-y = 0 $$

we have

$$ a (2 \lambda +3) x-\sqrt{2} \left(\sqrt{a^2 \left(\lambda ^2+4\right) x^2}+\sqrt{x^2 \left(a^2+(a-1)^2 \lambda ^2\right)}\right)-\lambda x=0 $$

or

$$ \left(a (2 \lambda +3)-\sqrt{2} \sigma(x)\left(\sqrt{a^2 \left(\lambda ^2+4\right)}+\sqrt{ \left(a^2+(a-1)^2 \lambda ^2\right)}\right)-\lambda \right)x=0 $$

where $\sigma(x)$ is the sign function. Now discarding the trivial solution $x=0$, for $\sigma(x)=1$ the solution is $\lambda=a=0$ and for $\sigma(x)=-1$ the solution is $\lambda =a=2$