What is ((x-3)^2)/4+((y-3)^2)/(16)=1 ?

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

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((x-3)^2)/4+((y-3)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

(h, k) = (3, 3), a = 2, b = 4

Solution steps

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 3), b = 4, a = 2

4 x^{2} + 2 y^{2} = 16

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

x^{2} + 4 y^{2} + 9 x + 9 y - 15 = 0

4 x^{2} + 16 y^{2} - 8 x + 96 y + 84 = 0

\frac{x ^{2}}{9} + \frac{2 x}{9} + y^{2} - 6 y + 9 = 0

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