What is the center ((x+2)^2}{25}+\frac{(y-4)^2)/4 =1 ?

The center ((x+2)^2}{25}+\frac{(y-4)^2)/4 =1 is (-2,4)

Popular >

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

center

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

(- 2, 4)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 4), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 4), a = 5, b = 2

And the center is:

(- 2, 4)

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

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

9 (x - 1)^{2} + 4 y^{2} = 36

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

x^{2} + 4 y^{2} - 10 x - 40 y + 121 = 0

What is the center ((x+2)^2}{25}+\frac{(y-4)^2)/4 =1 ?
The center ((x+2)^2}{25}+\frac{(y-4)^2)/4 =1 is (-2,4)