What is the center 25(x-5)^2+16(y+4)^2=400 ?

The center 25(x-5)^2+16(y+4)^2=400 is (5,-4)

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center 25(x-5)^2+16(y+4)^2=400

center

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

(5, - 4)

Solution steps

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

Rewrite

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(5, - 4)

4 x^{2} + 25 y^{2} - 32 x + 200 y + 364 = 0

(x 3)^{2} + (y + 4)^{2} = 25

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

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

What is the center 25(x-5)^2+16(y+4)^2=400 ?
The center 25(x-5)^2+16(y+4)^2=400 is (5,-4)