What is ((x-3)^2)/(64)+((y-3)^2)/(16)=1 ?

The solution to ((x-3)^2)/(64)+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(3,3),a=8,b=4

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((x-3)^2)/(64)+((y-3)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{16} = 1

(h, k) = (3, 3), a = 8, b = 4

Solution steps

\frac{( x - 3 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{8 ^{2}} + \frac{( y - 3 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 3), a = 8, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 3), a = 8, b = 4

4 x^{2} + y^{2} + 48 x + 128 = 0

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{16} = 1

(x - 1)^{2} + (2 y + 4)^{2} = 16

64 xx - 256 x + 81 yy + 810 y = 2903

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

What is ((x-3)^2)/(64)+((y-3)^2)/(16)=1 ?
The solution to ((x-3)^2)/(64)+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(3,3),a=8,b=4