Find the maximum of $a$ such that: $xy^2+4x^2y+5=0,x>0,ax

Question

Find the maximum of real number $a$ such that: $xy^2+4x^2y+5=0,x>0,ax

I think since $x>0$ so $y$ must be negative,but can't extract extra relation between $x,y$ from the given equation...

Answer 1

Obviously $x\ne0$ and $y\ne0$ for the equation to hold. With $t = \frac{y}{x} \ne 0$ the condition can be written as $a \lt t$ and the equation becomes:

$$ t^2 x^3 + 4 t x^3 + 5 = 0 \iff x^3 = \frac{-5}{t^2+4t}= \frac{-5}{t(t+4)} $$

Then $x \gt 0$ $\iff$ $x^3 \gt 0$ $\iff$ $t(t+4) \lt 0$ $\iff$ $t \in (-4,0)\,$ therefore $a \lt t$ for all $t$ iff $a \le -4$ so the maximum value of $a$ is $-4$.