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

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

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

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

(h, k) = (3, - 1), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 5, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 3, a = 5

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