What is ((x-3)^2}{16}+(\frac{(y+4))/9)^2=1 ?

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

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((x-3)^2}{16}+(\frac{(y+4))/9)^2=1

\frac{( x - 3 ) ^{2}}{16} + (\frac{( y + 4 )}{9})^{2} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{16} + (\frac{y + 4}{9})^{2} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + (\frac{y + 4}{9})^{2} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

9 x^{2} + 5 y^{2} = 180

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{29} = 1

\frac{( x - 4 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{10} = 1

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

\frac{( x - 5 ) ^{2}}{47} + \frac{( y + 3 ) ^{2}}{30} = 1

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