Consider the following two equations:
$$\sqrt {3x} \left(1 + \frac{1}{x+y} \right) = 2 $$
$$\sqrt{7y}\left(1 -\frac{1}{x+y}\right)= 4\sqrt{2}$$
. It is expected to verify if the equations possess a solution and if yes what is the floor of $ y/x $ for that solution.
. What have I tried:
. Eliminating the $\frac {1}{x+y}$ term from the two equations leads to : $$ \frac{2}{\sqrt{3x}} + \frac{4\sqrt{2}}{\sqrt{7y}} = 2 $$ Which gives $$ y = \frac{24x}{7(1- \sqrt{3x})^2}$$ Also from the first equation value of y is :
$$ y = \frac{\sqrt{3x}}{2-\sqrt{3x}} - x $$
Equating the two values of y and simplifying resulted in a polynomial with fractional powers of x. This proved to be a dead end for me.
. Note: some underlying inquisitions:
- Is there a method for simplifying such equations that I am unaware of?
- Are the given two equations representing hyperbola? (Checked the graph on desmos!). If yes how can one see that without referring to the plot.
No, the curves cannot be hyperbolas.
The first equation is necessarily a subset of the locus of points satisfying $3x(x+y+1)^2 = 4(x+y)^2$, and similarly for the second, $7y(x+y-1)^2 = 8(x+y)^2,$ which are plotted in the neighborhood of $[-1,1]^2$ below.
The first curve is tangent to the $y$-axis, and the second tangent to the $x$-axis (both easily verified through implicit differentiation). When zoomed out, we see:
So this suggests that the given equations cannot be sections of hyperbolae. They may be asymptotic to conic sections, but their degree is $3$, not $2$.