What is the center 4(x+5)^2+8(y-2)^2=32 ?

The center 4(x+5)^2+8(y-2)^2=32 is (-5,2)

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center 4(x+5)^2+8(y-2)^2=32

center

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

(- 5, 2)

Solution steps

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

Rewrite

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 5, 2), a = 22, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, 2), a = 22, b = 2

And the center is:

(- 5, 2)

4 x^{2} + 9 y^{2} + 24 x + 18 y + 9 = 0

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

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

x^{2} - 4 x + 9 y^{2} + 54 y + 4 = 0

What is the center 4(x+5)^2+8(y-2)^2=32 ?
The center 4(x+5)^2+8(y-2)^2=32 is (-5,2)