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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{x ^{2}}{7 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0

16 x^{2} + 25 y^{2} = 1600

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

4 x^{2} + 9 y^{2} - 64 x + 54 y + 301 = 0

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