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

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

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

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

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

Solution steps

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

Rewrite

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

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

6 x^{2} + 96 x + 16 y^{2} - 40 y = 315

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

4 x^{2} + y^{2} - 16 x + 6 y + 9 = 0

x^{2} + 9 y^{2} = 18

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