What is ((x-2)^2)/9+((y-3)^2)/5 =1 ?

The solution to ((x-2)^2)/9+((y-3)^2)/5 =1 is Ellipse with (h,k)=(2,3),a=3,b=sqrt(5)

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((x-2)^2)/9+((y-3)^2)/5 =1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

(h, k) = (2, 3), a = 3, b = 5

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 3), a = 3, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 3), a = 3, b = 5

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

16 x^{2} + 9 y^{2} - 128 x - 90 y + 337 = 0

(x + 6)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{16}} = 1

4 (x - 1)^{2} + 9 y^{2} = 36

\frac{( x ^{2} + 6 x + 9 )}{49} + \frac{( y ^{2} - 2 y + 1 )}{45} = 1

What is ((x-2)^2)/9+((y-3)^2)/5 =1 ?
The solution to ((x-2)^2)/9+((y-3)^2)/5 =1 is Ellipse with (h,k)=(2,3),a=3,b=sqrt(5)