What is ((x+3)^2)/9+((y-3)^2)/(16)=1 ?

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

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((x+3)^2)/9+((y-3)^2)/(16)=1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

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

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

1 x^{2} + 3 y^{2} + 9 x + 8 y - 14 = 0

25 x^{2} + 16 y^{2} + 100 x - 64 y - 236 = 0

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