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

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

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((x+2)^2)/9+((y+3)^2)/(25)=1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

x^{2} + y^{2} - 2 y > 0

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{C _{1}} = 1 - \frac{y ^{2}}{C _{2}}

25 x^{(2)} + 16 y^{(2)} - 96 y - 156 = 0

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} = 1

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