What is the center 4x^2+y^2-48x+4y+48=0 ?

The center 4x^2+y^2-48x+4y+48=0 is (6,-2)

Popular >

center 4x^2+y^2-48x+4y+48=0

center

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

(6, - 2)

Solution steps

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

Rewrite

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

in the form of the standard ellipse equation

\frac{( x - 6 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{1 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (6, - 2), a = 5, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (6, - 2), b = 10, a = 5

And the center is:

(6, - 2)

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

9 (x - 1)^{2} + 4 y^{2} = 36

9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0

x^{2} + 4 y^{2} - 10 x - 40 y + 121 = 0

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

What is the center 4x^2+y^2-48x+4y+48=0 ?
The center 4x^2+y^2-48x+4y+48=0 is (6,-2)