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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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