What is the center ((x-5)^2}{16}+\frac{(y+3)^2)/4 =1 ?

The center ((x-5)^2}{16}+\frac{(y+3)^2)/4 =1 is (5,-3)

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center ((x-5)^2}{16}+\frac{(y+3)^2)/4 =1

center

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{4} = 1

(5, - 3)

Solution steps

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), a = 4, b = 2

And the center is:

(5, - 3)

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 1 ) ^{2}}{9} = 1

2 x^{2} + y^{2} - 2 x + 16 y + 1 = 0

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

25 x^{2} + 16 y^{2} - 150 x + 128 y + 81 = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

What is the center ((x-5)^2}{16}+\frac{(y+3)^2)/4 =1 ?
The center ((x-5)^2}{16}+\frac{(y+3)^2)/4 =1 is (5,-3)