What is the center (x^2)/7-(7y^2)/(42)=1 ?

The center (x^2)/7-(7y^2)/(42)=1 is (0,0)

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center (x^2)/7-(7y^2)/(42)=1

center

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

Rewrite

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{( 7 ) ^{2}} - \frac{( y - 0 ) ^{2}}{( 6 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 7, b = 6

And the center is:

(0, 0)

4 y^{2} - 49 x^{2} - 392 x + 16 y - 964 = 0

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

x^{2} - y^{2} = 16

\frac{x ^{2}}{4} - \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{144} - \frac{y ^{2}}{25} = 1

What is the center (x^2)/7-(7y^2)/(42)=1 ?
The center (x^2)/7-(7y^2)/(42)=1 is (0,0)