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

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

Popular >

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

center

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

(1, - 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), b = 4, a = 2

And the center is:

(1, - 2)

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1