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

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

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

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

(h, k) = (2, - 1), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), a = 5, b = 3

4 x^{2} + 5 y^{2} = 20

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{81} = 1

36 x^{2} + 16 y^{2} - 180 x + 96 y - 207 = 0

x^{2} + 3 y^{2} = 9

12 x^{2} + 16 y^{2} - 120 x + 64 y + 172 = 0

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