What is ((x+2)^2)/9+((y+3)^2)/(16)=1 ?

The solution to ((x+2)^2)/9+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(-2,-3),a=3,b=4

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((x+2)^2)/9+((y+3)^2)/(16)=1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

\frac{1}{2} (3 x^{2} + y^{2}) - 2 (x - y) = 0

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 11

9 x^{2} + 25 y^{2} + 18 x + 50 y - 191 = 0

What is ((x+2)^2)/9+((y+3)^2)/(16)=1 ?
The solution to ((x+2)^2)/9+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(-2,-3),a=3,b=4