Question

what is the equation in standard form? (4x)^2+y^2-48x-4y+48=0

Answer 1

\[4x^2+y^2-48x-4y+48=0\] is an ellipse
you want the standard form of this?

Answer 2

\[4x^2-48x+y^2-4y=-48\]
\[4(x^2-12x)+y^2-4y=-48\]
\[4(x^2-12x+36)+y^2-4y+4=-48+144+4=100\]
\[4(x-6)^2+(y-2)^2=100\]

Answer 3

divide by 100 to get
\[\frac{(x-6)^2}{25}+\frac{(y-2)^2}{100}=1\]

Answer 4

thank you!!!