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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

7 x^{2} + 18 y^{2} = 63

\frac{( x + 3 ) ^{2}}{100} + \frac{( y + 4 ) ^{2}}{32} = 1

\frac{x ^{2}}{7} + \frac{y ^{2}}{4} = 1

\frac{( x - 2 ) ^{2}}{5} + \frac{( y + 3 ) ^{2}}{9} = 1

4 x^{2} + 9 y^{2} - 24 x + 18 y + 9 = 0

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