What is the center ((x-4)^2)/(100)+((y+5)^2)/(64)=1 ?

The center ((x-4)^2)/(100)+((y+5)^2)/(64)=1 is (4,-5)

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center ((x-4)^2)/(100)+((y+5)^2)/(64)=1

center

\frac{( x - 4 ) ^{2}}{100} + \frac{( y + 5 ) ^{2}}{64} = 1

(4, - 5)

Solution steps

\frac{( x - 4 ) ^{2}}{100} + \frac{( y + 5 ) ^{2}}{64} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{100} + \frac{( y + 5 ) ^{2}}{64} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{1 0 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, - 5), a = 10, b = 8

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, - 5), a = 10, b = 8

And the center is:

(4, - 5)

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

2 x^{2} - 12 x + y^{2} + 6 y + 26 = 0

9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

What is the center ((x-4)^2)/(100)+((y+5)^2)/(64)=1 ?
The center ((x-4)^2)/(100)+((y+5)^2)/(64)=1 is (4,-5)