What is the center 4x^2+y^2+8x-4y+4=0 ?

The center 4x^2+y^2+8x-4y+4=0 is (-1,2)

Popular >

center 4x^2+y^2+8x-4y+4=0

center

4 x^{2} + y^{2} + 8 x - 4 y + 4 = 0

(- 1, 2)

Solution steps

4 x^{2} + y^{2} + 8 x - 4 y + 4 = 0

Rewrite

4 x^{2} + y^{2} + 8 x - 4 y + 4 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{1 ^{2}} + \frac{( y - 2 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 2), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 2), b = 2, a = 1

And the center is:

(- 1, 2)

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

\frac{x ^{2}}{16} + (y - 4)^{2} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1