Rotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form.a. xy + 1 = 0

Answer:[tex]\dfrac{x'^2}{2}-\dfrac{y'^2}{2}=1[/tex]Step-by-step explanation:The rotation by angle [tex]\theta[/tex] formulas are[tex]\left\{\begin{array}{l}x=x'\cos \theta-y'\sin \theta\\y=x'\sin \theta+y' \cos \theta\end{array} \right.[/tex]To eliminate the xy-term, we have to rotate by 45°, so[tex]\left\{\begin{array}{l}x=x'\cos 45^{\circ}-y'\sin 45^{\circ}\\y=x'\sin 45^{\circ} +y' \cos 45^{\circ}\end{array} \right.[/tex][tex]\left\{\begin{array}{l}x=x'\dfrac{\sqrt{2}}{2}-y'\dfrac{\sqrt{2}}{2}\\y=x'\dfrac{\sqrt{2}}{2} +y' \dfrac{\sqrt{2}}{2}\end{array} \right.[/tex]Substitute them into the equation [tex]xy+1=0:[/tex][tex]\left(x'\dfrac{\sqrt{2}}{2}-y'\dfrac{\sqrt{2}}{2}\right)\cdot \left(x'\dfrac{\sqrt{2}}{2}+y'\dfrac{\sqrt{2}}{2}\right)+1=0\\ \\\left(x'\dfrac{\sqrt{2}}{2}\right)^2 -\left(y'\dfrac{\sqrt{2}}{2}\right)^2+1=0\\ \\\dfrac{x'^2}{2}-\dfrac{y'^2}{2}=1[/tex]