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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

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

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

16 x^{2} - 64 x + 25 y^{2} + 150 y = 111

\frac{x ^{2}}{100} + \frac{y ^{2}}{25} = 1

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