What is ((x-5)^2)/9+((y+3)^2)/4 =1 ?

The solution to ((x-5)^2)/9+((y+3)^2)/4 =1 is Ellipse with (h,k)=(5,-3),a=3,b=2

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((x-5)^2)/9+((y+3)^2)/4 =1

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{4} = 1

(h, k) = (5, - 3), a = 3, b = 2

Solution steps

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 3, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), a = 3, b = 2

\frac{y ^{2}}{16} + \frac{x ^{2}}{36} = 1

\frac{( x + 2 ) ^{2}}{55} + \frac{( y - 2 ) ^{2}}{64} = 1

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{4} = 1

- 8 x^{2} - y^{2} - 4 x + 5 y + 9 = 0

(\frac{X}{3})^{2} + (\frac{Y}{2})^{2} = 1

What is ((x-5)^2)/9+((y+3)^2)/4 =1 ?
The solution to ((x-5)^2)/9+((y+3)^2)/4 =1 is Ellipse with (h,k)=(5,-3),a=3,b=2