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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

3 x^{2} + y^{2} = 6

0.2 x^{2} + 2 y^{2} + 4 x - 8 y - 12 = 0

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

5 x^{2} + 4 y^{2} = 1

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

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