What is the center (x+4)^2+((y-3)/4)^2=1 ?

The center (x+4)^2+((y-3)/4)^2=1 is (-4,3)

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center (x+4)^2+((y-3)/4)^2=1

center

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

(- 4, 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(- 4, 3)

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

\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

What is the center (x+4)^2+((y-3)/4)^2=1 ?
The center (x+4)^2+((y-3)/4)^2=1 is (-4,3)