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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{100} + \frac{y ^{2}}{4} = 1

\frac{( x - 6 ) ^{2}}{100} + \frac{( y + 3 ) ^{2}}{16} = 1

2 x^{2} + 4 y^{2} + 9 x + 9 y - 17 = 0

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

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

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