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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

4 x^{2} + 25 y^{2} - 24 x + 250 y + 561 = 0

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

100 x^{2} + 64 y^{2} = 1600

\frac{x ^{2}}{1} + \frac{y ^{2}}{9} = 1

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

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