What is the center (x^2)/7+(y^2)/(16)=1 ?

The center (x^2)/7+(y^2)/(16)=1 is (0,0)

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center (x^2)/7+(y^2)/(16)=1

center

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

Rewrite

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( 7 ) ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 7, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 4, a = 7

And the center is:

(0, 0)

25 (x - 2)^{2} + 9 (y + 3)^{2} = 225

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

\frac{( x + 5 ) ^{2}}{5} + \frac{( y + 3 ) ^{2}}{3} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{121} = 1

9 x^{2} + 4 y^{2} - 18 x - 16 y - 11 = 0

What is the center (x^2)/7+(y^2)/(16)=1 ?
The center (x^2)/7+(y^2)/(16)=1 is (0,0)