Question

Trace the following central conics.
$3(2x-3y+4{)}^2+2(3x+2y-5{)}^2=78.$

Answer 1

$3{(2x-3y+4)}^2+2{(3x+2y-5)}^2=783\times 13{\left(\frac{2x-3y+4}{\sqrt{13}}\right)}^2+2\times 13{\left(\frac{3x+2y-5}{\sqrt{13}}\right)}^2=78X=\frac{2x-3y+4}{\sqrt{13}},Y=\frac{3x+2y-5}{\sqrt{13}}39X^2+26Y^2=78\frac{X^2}{2}+\frac{Y^2}{3}=1$

Comparing with the standard equation of ellipse

$a=\sqrt{2},b=\sqrt{3}$