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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

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

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

\frac{x ^{2}}{256} + \frac{y ^{2}}{400} = 2

3 x^{2} + 4 y^{2} + 6 x - 24 y + 27 = 0

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