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

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

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

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

(h, k) = (3, - 1), a = 3, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 4, a = 3

\frac{x ^{2}}{580 0 ^{2}} + \frac{y ^{2}}{150 0 ^{2}} = 1

9 x^{2} + y^{2} - 54 x - 6 y + 81 = 0

x^{2} = \frac{4 - y ^{2}}{2}

64 x^{2} - 192 x - 156 + 100 y^{2} - 200 y = 0

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{81} = 1

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