Question

Eliminate $x,y,z$ from the equations $yz=a^2,zx=b^2,xy=c^2,x^2+y^2+z^2=d^2$.

Answer 1

Given equations are $yz=a^2,zx=b^2,xy=c^2,x^2+y^2+z^2=d^2$

$\Rightarrow xyz=a^{2}x$, $yxz=b^{2}y$, $xyz=c^{2}z$
We can call all these equal to some constant k
i.e $a^{2}x=b^{2}y=c^{2}z=k$
Therefore,
$x=\frac{k}{a^2},y=\frac{k}{b^2},z=\frac{k}{c^2}$
Given $x^2+y^2+z^2=d^2$
Substitute $x,y$ and $z$ in this equation, we get
$\frac{k^2}{a^4}+\frac{k^2}{b^4}+\frac{k^2}{c^4}=d^2$
$k^2\left(\frac{a^4{b}^4+c^4{a}^4+b^4{c}^4}{a^4{b}^4{c}^4}\right)=d^2$
$k=\frac{d{a}^2{b}^2{c}^2}{\sqrt{a^4{b}^4+c^4{a}^4+b^4{c}^4}}$
We know that, $b^{2}y=k$
Therefore, $y=\frac{d{a}^2{c}^2}{\sqrt{a^4{b}^4+c^4{a}^4+b^4{c}^4}}$
and $z=\frac{d{a}^2{b}^2}{\sqrt{a^4{b}^4+c^4{a}^4+b^4{c}^4}}$
Also, $yz=a^2$
$\therefore \frac{d^2{a}^4{b}^2{c}^2}{a^4{b}^4+c^4{b}^4+a^4{c}^4}=a^2$
The eliminated equation is
$d^2{a}^2{b}^2{c}^2=a^4{b}^4+c^4{b}^4+a^4{c}^4$