What is ((x-2)^2)/4+((y-3)^2)/9 =1 ?

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

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((x-2)^2)/4+((y-3)^2)/9 =1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

x^{2} + 2 y^{2} - 10 x + 8 y + 29 = 0

16 x^{2} + 4 y^{2} - 32 x + 16 y - 32 = 0

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

\frac{( x - 5 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{81} = 1

x^{2} + 9 y^{2} \leq 16

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