What is ((x+3)^2}{25}+\frac{(y+1)^2)/9 =1 ?

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

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((x+3)^2}{25}+\frac{(y+1)^2)/9 =1

\frac{( x + 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

4 x^{2} - 48 x + 9 y^{2} + 72 y + 144 = 0

\frac{( x - 2.22 ) ^{2}}{2.54} + \frac{y ^{2}}{1.11} = 2

25 x^{2} + 16 y^{2} - 100 x - 64 y - 236 = 0

12 x - 5 y + 10 x^{2} + 25 = \frac{- ( 100 y ^{2} )}{10 + 3}

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{36} = 1

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