Question

Уравнение поверхности к каноническому виду: 2x^2 + 2y^2 + 2z^2 + 4y - 3z + 2 = 0

Answer 1

$2 x^{2} +2(y^2+2y+1)-2+2(z^2- \frac{3}{2}z+ \frac{9}{16} )- \frac{9}{16} +2=0 \\ 2 x^{2} +2 (y+1)^2+2(z- \frac{3}{4} )^2= \frac{9}{16} \\ \frac{x^2}{ (\frac{3}{(4 \sqrt{2} )})^2 } + \frac{(y+1)^2}{ (\frac{3}{4 \sqrt{2}})^2 } + \frac{(z- \frac{3}{4})^2 }{ (\frac{3}{(4 \sqrt{2} )})^2 } =1$