What is ((x+3)^2)/4+((y+2)^2)/(16)=1 ?

The solution to ((x+3)^2)/4+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(-3,-2),a=2,b=4

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((x+3)^2)/4+((y+2)^2)/(16)=1

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

(h, k) = (- 3, - 2), a = 2, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, - 2), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, - 2), b = 4, a = 2

\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1

\frac{- x ^{2}}{2 ^{2}} - \frac{y ^{2}}{4 ^{2}} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 8 ) ^{2}}{16} = 1

36 (x + 2)^{2} + (y - 4)^{2} = 72

\frac{( x - 2 ) ^{2}}{4} + \frac{y ^{2}}{33} = 1

What is ((x+3)^2)/4+((y+2)^2)/(16)=1 ?
The solution to ((x+3)^2)/4+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(-3,-2),a=2,b=4