Cfrgtkky

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 $-y^2=aiff y^2 = -a$. 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 $bneq 0$ and therefore from $II$ we get $y=frac{b}{2x}$, substituting with $I Longrightarrow x^2-(frac{b}{2x})^2=aiff 4x^4 - b^2 = 4x^2a iff x^4 -ax^2 - frac{b^2}{4}=0 (*)$

Completing the square

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

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

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

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:

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

Thank you for your time.

Actually, the statement is true even if $b=0$: just take $(x,y)=left(pmsqrt a,0right)$ if $ageqslant0$ and $(x,y)=left(0,pmsqrt{-a}right)$.

Otherwise, your approach is fine, but not your computations. You should have obtained, when $bneq0$,$$pmleft(sqrt{frac{a+sqrt{a^2+b^2}}2},frac b{|b|}sqrt{frac{-a+sqrt{a^2+b^2}}2}right),$$where, of course,$$frac b{lvert brvert}=begin{cases}1&text{ if }b>0\-1&text{ otherwise.}end{cases}$$

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

Suppose $age0$; 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 $bne0$, 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
$$
x=pmsqrt{frac{sqrt{a^2+b^2}+a}{2}}
$$
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
$$
y=pmsqrt{frac{sqrt{a^2+b^2}-a}{2}}
$$
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 $b<0$.

Geometric solution

If $;a,bneq 0;$ 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.