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

The center ((x-4)^2)/4+(y^2)/(16)=1 is (4,0)

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

center

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

(4, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 0), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 0), b = 4, a = 2

And the center is:

(4, 0)

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

\frac{( x - 4 ) ^{2}}{100} + \frac{( y + 5 ) ^{2}}{64} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

2 x^{2} - 12 x + y^{2} + 6 y + 26 = 0

What is the center ((x-4)^2)/4+(y^2)/(16)=1 ?
The center ((x-4)^2)/4+(y^2)/(16)=1 is (4,0)