What is the center 16(x-1)^2+9(y+2)^2=144 ?

The center 16(x-1)^2+9(y+2)^2=144 is (1,-2)

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center 16(x-1)^2+9(y+2)^2=144

center

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

(1, - 2)

Solution steps

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

Rewrite

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{3 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), b = 4, a = 3

And the center is:

(1, - 2)

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1

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

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

2 x^{2} + 2 y^{2} + 16 y = 0

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

What is the center 16(x-1)^2+9(y+2)^2=144 ?
The center 16(x-1)^2+9(y+2)^2=144 is (1,-2)