What is the center ((x+3)^2)/(49)+(y^2)/(36)=1 ?

The center ((x+3)^2)/(49)+(y^2)/(36)=1 is (-3,0)

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center ((x+3)^2)/(49)+(y^2)/(36)=1

center

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

(- 3, 0)

Solution steps

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{7 ^{2}} + \frac{( y - 0 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 0), a = 7, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 0), a = 7, b = 6

And the center is:

(- 3, 0)

\frac{( x + 2 ) ^{2}}{81} + \frac{y ^{2}}{36} = 1

25 x^{2} + 16 y^{2} < 400

16 x^{2} + 4 y^{2} - 64 = 0

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

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

What is the center ((x+3)^2)/(49)+(y^2)/(36)=1 ?
The center ((x+3)^2)/(49)+(y^2)/(36)=1 is (-3,0)