What is ((x-2)^2)/9+((y+1)^2)/(36)=1 ?

The solution to ((x-2)^2)/9+((y+1)^2)/(36)=1 is Ellipse with (h,k)=(2,-1),a=3,b=6

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((x-2)^2)/9+((y+1)^2)/(36)=1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1

(h, k) = (2, - 1), a = 3, b = 6

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{3 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), b = 6, a = 3

x^{2} + 4 y^{2} - 64 = 0

\frac{x ^{2}}{5} + \frac{y ^{2}}{8} = 1

\frac{x ^{2}}{140} + \frac{( y - 6 ) ^{2}}{4} = 1

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

9 x^{2} - 54 x + 4 y^{2} = - 45

What is ((x-2)^2)/9+((y+1)^2)/(36)=1 ?
The solution to ((x-2)^2)/9+((y+1)^2)/(36)=1 is Ellipse with (h,k)=(2,-1),a=3,b=6