What is the center ((x-2)^2)/(64)+((y+4)^2)/(16)=1 ?

The center ((x-2)^2)/(64)+((y+4)^2)/(16)=1 is (2,-4)

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center ((x-2)^2)/(64)+((y+4)^2)/(16)=1

center

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

(2, - 4)

Solution steps

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{8 ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 4), a = 8, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 4), a = 8, b = 4

And the center is:

(2, - 4)

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

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

x^{2} + 10 x + 9 y^{2} + 16 = 0

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

What is the center ((x-2)^2)/(64)+((y+4)^2)/(16)=1 ?
The center ((x-2)^2)/(64)+((y+4)^2)/(16)=1 is (2,-4)