Question

How do you get 3x^2+7y^2+24x-70y=-202
into the form (x+4)^2/7+(y-5)^2/3=1?

Answer 1

\[3x^2+24x+7y^2-70y=-202\]
\[3(x^2+8x)+7(y^2-10y)=-202\]
\[3(x^2+8x+(\frac{8}{2})^2)+7(y^2-10x+(\frac{10}{2})^2)=-202+3(\frac{8}{2})^2+7(\frac{10}{2})^2\]
\[3(x^2+8x+4^2)+7(y^2-10x+5^2)=-202+3(16)+7(25)\]
\[3(x+4)^2+7(y-5)^2=-202+48+175\]
\[3(x+4)^2+7(y-5)^2=21\]
\[\frac{3}{21}(x+4)^2+\frac{7}{21}(y-5)^2=\frac{21}{21}\]
\[\frac{1}{7}(x+4)^2+\frac{1}{3}(y-5)^2=1\]

Answer 2

completing the square dave is all you really need to do this trick

Answer 3

Ah, I see. I totally missed that. Thanks!

Answer 4

np