What is ((x-3)^2)/(36)+((y-3)^2)/(32)=1 ?

The solution to ((x-3)^2)/(36)+((y-3)^2)/(32)=1 is Ellipse with (h,k)=(3,3),a=6,b=4sqrt(2)

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((x-3)^2)/(36)+((y-3)^2)/(32)=1

\frac{( x - 3 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{32} = 1

(h, k) = (3, 3), a = 6, b = 42

Solution steps

\frac{( x - 3 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{32} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{32} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{6 ^{2}} + \frac{( y - 3 ) ^{2}}{( 4 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 3), a = 6, b = 42

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 3), a = 6, b = 42

x^{2} + 4 y^{2} - 20 x + 84 = 0

(x^{2} + y^{2} - 16 x + 8 y + 80) = 4 (x^{2} + y^{2} - 4 x - 8 y + 20)

16 (x - 2)^{2} + 25 (y + 3)^{2} = 124

3 x^{2} + y^{2} + 12 x - 2 y + 1 = 0

10 y^{2} + x^{2} = 5

What is ((x-3)^2)/(36)+((y-3)^2)/(32)=1 ?
The solution to ((x-3)^2)/(36)+((y-3)^2)/(32)=1 is Ellipse with (h,k)=(3,3),a=6,b=4sqrt(2)