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)

((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

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

36 x^{2} + 9 y^{2} + 288 x - 90 y + 477 = 0

(\frac{x ^{2}}{72}) + (\frac{y ^{2}}{21}) = 1

16 x^{2} + 25 y^{2} = 1400

5 x^{2} + 9 y^{2} - 80 x + 54 y + 221 = 0

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)