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

(x^{2} + 9 x) + (3 y^{2} + 10 y) = 2

\frac{x ^{2}}{100} + \frac{y ^{2}}{50} = 1

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

(2 x + 3)^{2} + (y - 2)^{2} = 26

4 x^{2} + y^{2} - 32 x + 4 y + 64 = 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