What is ((x-2)^2)/9+((y+2)^2)/(25)=1 ?

The solution to ((x-2)^2)/9+((y+2)^2)/(25)=1 is Ellipse with (h,k)=(2,-2),a=3,b=5

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((x-2)^2)/9+((y+2)^2)/(25)=1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

(h, k) = (2, - 2), a = 3, b = 5

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 2), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 2), b = 5, a = 3

x^{2} + 4 y^{2} + 10 x + 8 y - 13 = 0

25 x^{2} + 4 y^{2} - 100 x + 8 y + 91 = 0

4 x^{2} + 9 y^{2} - 54 y + 45 = 0

x^{2} + 4 y^{2} = 5

3 x^{2} + 4 y^{2} + 18 x + 8 y + 19 = 0

What is ((x-2)^2)/9+((y+2)^2)/(25)=1 ?
The solution to ((x-2)^2)/9+((y+2)^2)/(25)=1 is Ellipse with (h,k)=(2,-2),a=3,b=5