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

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

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

center

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

(1, - 5)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 5), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 5), a = 4, b = 2

And the center is:

(1, - 5)

x^{2} + y^{2} - 6 + 2 y + 6 = 0

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

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

x^{2} + y^{2} - 12 y = 0

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

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