What is the center 13x^2+49y^2=637 ?

The center 13x^2+49y^2=637 is (0,0)

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center 13x^2+49y^2=637

center

13 x^{2} + 49 y^{2} = 637

(0, 0)

Solution steps

13 x^{2} + 49 y^{2} = 637

Rewrite

13 x^{2} + 49 y^{2} = 637

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{7 ^{2}} + \frac{( y - 0 ) ^{2}}{( 13 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 7, b = 13

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 7, b = 13

And the center is:

(0, 0)

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

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