What is the center (x^2)/(25)+(y^2)/(16)=1 ?

The center (x^2)/(25)+(y^2)/(16)=1 is (0,0)

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center (x^2)/(25)+(y^2)/(16)=1

center

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

Rewrite

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{5 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 5, b = 4

And the center is:

(0, 0)

3 x^{2} + 3 y^{2} = 75

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

y^{2} = 12 x

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

x^{2} + y^{2} + 4 y = 0

What is the center (x^2)/(25)+(y^2)/(16)=1 ?
The center (x^2)/(25)+(y^2)/(16)=1 is (0,0)