What is ((x+3)^2)/4+((y+1)^2)/9 =1 ?

The solution to ((x+3)^2)/4+((y+1)^2)/9 =1 is Ellipse with (h,k)=(-3,-1),a=2,b=3

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((x+3)^2)/4+((y+1)^2)/9 =1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

1 = 4 x^{2} + 2 y^{2}

x^{2} + 2 y^{2} - 6 y = 18

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

\frac{( x ^{2} )}{63500} + \frac{( y ^{2} )}{50900} = 1

\frac{( x + 5 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{4} = 1

What is ((x+3)^2)/4+((y+1)^2)/9 =1 ?
The solution to ((x+3)^2)/4+((y+1)^2)/9 =1 is Ellipse with (h,k)=(-3,-1),a=2,b=3