Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$

Question

The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no solution for the equation System, vice versa if $y=0$ and $a<0$.

For the general case I have assumed that $b\neq 0$ and therefore from $II$ we get $y=\frac{b}{2x}$, substituting with $I \Longrightarrow x^2-(\frac{b}{2x})^2=a\iff 4x^4 - b^2 = 4x^2a \iff x^4 -ax^2 - \frac{b^2}{4}=0 (*)$

Completing the square

$(*)\iff x^4 - 2\frac{ax^2}{2}-\frac{b^2}{4} \iff x^4 - 2\frac{ax^2}{2} +\frac{a^2}{4}-\frac{b^2}{4}-\frac{a^2}{4}\iff (x^2 - \frac{a}{2})^2+\frac{-b^2-a^2}{4}=0$

$\Rightarrow x^2-\frac{a}{2}=\sqrt{\frac{b^2+a^2}{4}} \Rightarrow x^2=\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2} \Rightarrow x = \pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}}$

$II\Longrightarrow y= \frac{b}{2(\pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}})}$

Can somebody tell my whether this Solutions are Right or not because when I have tried to verify it I end up with that term:

$\frac{5a^2+a(5\sqrt\frac{b^2+a^2}{4})(4\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2})}{4}$ which should be equal to $a$

Thank you for your time.

Answer 1

Actually, the statement is true even if $b=0$: just take $(x,y)=\left(\pm\sqrt a,0\right)$ if $a\geqslant0$ and $(x,y)=\left(0,\pm\sqrt{-a}\right)$.

Otherwise, your approach is fine, but not your computations. You should have obtained, when $b\neq0$,$$\pm\left(\sqrt{\frac{a+\sqrt{a^2+b^2}}2},\frac b{|b|}\sqrt{\frac{-a+\sqrt{a^2+b^2}}2}\right),$$where, of course,$$\frac b{\lvert b\rvert}=\begin{cases}1&\text{ if }b>0\\-1&\text{ otherwise.}\end{cases}$$

Answer 2

If $b=0$, then either $x$ or $y$ is zero.

Suppose $a\ge0$; then $x^2-y^2=a$ has the solution $x=\sqrt{a}$, $y=0$.

If $a<0$, then $x^2-y^2=a$ has the solution $x=0$, $y=\sqrt{-a}$.

In both cases, the constraint $xy=0$ is satisfied.

For the case $b\ne0$, you can substitute $y=b/(2x)$ and get the equation $$ 4x^4-4ax^2-b^2=0 $$ which is a biquadratic; the associated equation $4z^2-4az-b^2=0$ has a positive root, because $-b^2<0$, so also the biquadratic has a solution (actually two).

Your computations are wrong. The positive root of $4z^2-4az-b^2=0$ is $$ \frac{4a+\sqrt{16a^2+16b^2}}{8}=\frac{a+\sqrt{a^2+b^2}}{2} $$ Thus $$ x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} $$ From $y=b/(2x)$ we get \begin{align} y^2=\frac{b^2}{4x^2} &=\frac{b^2}{2(\sqrt{a^2+b^2}+a)} \\[6px] &=\frac{b^2}{2(\sqrt{a^2+b^2}+a)}\frac{\sqrt{a^2+b^2}-a}{\sqrt{a^2+b^2}-a} \\[6px] &=\frac{b^2(\sqrt{a^2+b^2}-a)}{2(a^2+b^2-a^2)}\\[6px] &=\frac{\sqrt{a^2+b^2}-a}{2} \end{align}

Therefore $$ y=\pm\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}} $$ The signs are not arbitrarily chosen: if you take the $+$ for $x$, then you will need to take the $+$ or the $-$ for $y$, according whether $b>0$ or $b<0$.

Answer 3

Geometric solution

If $\;a,b\neq 0\;$ the equations define two hyperbolas centered both in $0,$ where

• $I:x^2-y^2=a\;$ has asymptotes $\;y=\pm x,$ and lies in the left and right sectors ($a>0$) or in upper and lower sectors ($a<0$) delimited by the asymptotes;

• $II: 2xy= b\;$ has asymptotes $\;x=0,y=0\;$ and lies in the first and third quadrants ($b>0$) or in the second and fourth if $b<0$ (the quadrants are sectors delimited by asymptotes).

If $a=0$ or $b=0,$ the corresponding equation defines the two asymptotes of $I$ or of $II.$

Thus the two objects $I$ and $II$ necessarilly intersect in $2$ points if at least one of them is a hyperbola, and in one point in the degenerate case $a=0=b.$ The coordinates $(x,y)$ of these points are solutions of the system.