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

4 x^{2} + y^{2} - 24 x - 8 y + 8 = 0

1 = x^{2} + 3 y^{2}

4 x^{2} + y^{2} - 24 x + 2 y + 21 = 0

x^{2} + 5 y^{2} = \frac{1}{5}

9 x^{2} + 4 y^{2} + 18 x - 40 y + 73 = 0

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