What is the center (x^2)/(16)+(y^2)/(12)=1 ?

The center (x^2)/(16)+(y^2)/(12)=1 is (0,0)

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center (x^2)/(16)+(y^2)/(12)=1

center

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

Rewrite

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{( 2 3 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 4, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 4, b = 23

And the center is:

(0, 0)

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

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

36 x^{2} + 25 y^{2} = 900

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

What is the center (x^2)/(16)+(y^2)/(12)=1 ?
The center (x^2)/(16)+(y^2)/(12)=1 is (0,0)