What is the center ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?

The center ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (3,-5)

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center ((x-3)^2)/(49)+((y+5)^2)/(25)=1

center

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

(3, - 5)

Solution steps

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{7 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 5), a = 7, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 5), a = 7, b = 5

And the center is:

(3, - 5)

64 x^{2} + 100 y^{2} + 768 x - 600 y - 3196 = 0

2 x^{2} + 3 y^{2} - 4 x + 6 y - 1 = 0

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

4 x^{2} + 25 y^{2} = 100

2 (x + \frac{1}{4})^{2} + y^{2} = 8

What is the center ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?
The center ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (3,-5)