What is the center ((x+2)^2)/4+((y-4)^2)/(16)=1 ?

The center ((x+2)^2)/4+((y-4)^2)/(16)=1 is (-2,4)

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center ((x+2)^2)/4+((y-4)^2)/(16)=1

center

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

(- 2, 4)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(- 2, 4)

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

4 x^{2} + 9 y^{2} = 36

25 x^{2} + 9 y^{2} = 225

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

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

What is the center ((x+2)^2)/4+((y-4)^2)/(16)=1 ?
The center ((x+2)^2)/4+((y-4)^2)/(16)=1 is (-2,4)