What is ((x+3)^2}{16}+\frac{(y+2)^2)/9 =1 ?

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

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((x+3)^2}{16}+\frac{(y+2)^2)/9 =1

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

16 x^{(2)} + 25 y^{(2)} - 128 x + 100 y = 0

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 8 ) ^{2}}{36} = 1

16 x^{2} + 25 y^{2} = \frac{169}{4}

\frac{( x - 3 ) ^{2}}{16} + (\frac{( y + 4 )}{9})^{2} = 1

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