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

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

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

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

(h, k) = (- 3, 1), a = 4, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, 1), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 1), a = 4, b = 2

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

\frac{( x + 8 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{81} = 1

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1

3 x^{2} + 6 y^{2} - 4 x + 6 y - 3 = 0

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

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