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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

9 x^{2} + 4 y^{2} - 36 x - 8 y + 4 = 0

x^{2} + 3 y^{2} + 18 y + 18 = 0

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

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{36} = 1

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