What is the center ((x-10)^2)/4+(y^2)/(16)=1 ?

The center ((x-10)^2)/4+(y^2)/(16)=1 is (10,0)

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center ((x-10)^2)/4+(y^2)/(16)=1

center

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

(10, 0)

Solution steps

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

Rewrite

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 10 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (10, 0), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (10, 0), b = 4, a = 2

And the center is:

(10, 0)

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

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

25 x^{2} + 16 y^{2} + 100 x - 96 y = 156

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

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

What is the center ((x-10)^2)/4+(y^2)/(16)=1 ?
The center ((x-10)^2)/4+(y^2)/(16)=1 is (10,0)