What is the center ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?

The center ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is (1,-3)

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center ((x-1)^2)/(25)+((y+3)^2)/(5100)=1

center

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

(1, - 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{( 10 51 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 5, b = 1051

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), b = 1051, a = 5

And the center is:

(1, - 3)

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

2 x^{2} + 4 x = 19 - 3 y^{2}

9 x^{2} + 49 y^{2} + 18 x + 294 y - 9 = 0

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

What is the center ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?
The center ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is (1,-3)