What is ((x+2)^2}{16}+\frac{(y-2)^2)/9 =1 ?

The solution to ((x+2)^2}{16}+\frac{(y-2)^2)/9 =1 is Ellipse with (h,k)=(-2,2),a=4,b=3

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((x+2)^2}{16}+\frac{(y-2)^2)/9 =1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

9 x^{(2)} + 4 y^{(2)} + 36 x - 24 y + 36 = 0

9 (y^{2} + 6 y + 9) + 16 (x^{2} - 4 x + 4) = 81 + 64

\frac{( x - h ) ^{2}}{b ^{2}} + \frac{( y - k ) ^{2}}{a ^{2}} = 1

25 x^{(2)} + 9 y^{(2)} + 50 x - 108 y + 124 = 0

What is ((x+2)^2}{16}+\frac{(y-2)^2)/9 =1 ?
The solution to ((x+2)^2}{16}+\frac{(y-2)^2)/9 =1 is Ellipse with (h,k)=(-2,2),a=4,b=3