What is the center ((x+1)^2)/(64)+((y-1)^2)/(100)=1 ?

The center ((x+1)^2)/(64)+((y-1)^2)/(100)=1 is (-1,1)

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center ((x+1)^2)/(64)+((y-1)^2)/(100)=1

center

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

(- 1, 1)

Solution steps

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{8 ^{2}} + \frac{( y - 1 ) ^{2}}{1 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 1), a = 8, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 1), b = 10, a = 8

And the center is:

(- 1, 1)

36 x^{2} + 25 y^{2} = 900

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{4} = 1

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

What is the center ((x+1)^2)/(64)+((y-1)^2)/(100)=1 ?
The center ((x+1)^2)/(64)+((y-1)^2)/(100)=1 is (-1,1)