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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

5 x^{2} + 9 y^{2} = 45

9 x^{2} + y^{2} = 81

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

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

9 x^{2} + 4 y^{2} + 18 x - 16 y - 11 = 0

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