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

25 x^{2} + 16 y^{2} + 100 x - 96 y = 156

x^{2} + 9 y^{2} - 8 x + 36 y + 4 = 0

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

9 x^{2} - 54 x + y^{2} - 2 y = - 81

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{81} = 1

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