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

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

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

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

(h, k) = (3, - 2), a = 4, b = 9

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 4, b = 9

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), b = 9, a = 4

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

9 x^{2} + 16 y^{2} + 18 x - 64 y - 71 = 0

\frac{( x - 2.37 ) ^{2}}{2.85} + \frac{y ^{2}}{4.14} = 2

\frac{x ^{2}}{4} + \frac{y ^{2}}{8} = \frac{3}{4}

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

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