What is ((x-2)^2)/9+((y-3)^2)/(25)=1 ?

The solution to ((x-2)^2)/9+((y-3)^2)/(25)=1 is Ellipse with (h,k)=(2,3),a=3,b=5

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((x-2)^2)/9+((y-3)^2)/(25)=1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

9 x^{2} + 4 y^{2} + 36 x + 24 y + 36 = 0

x^{2} + 4 y^{2} - 80 y + 143 = 0

9 x^{2} + 8 y^{2} - 54 x - 16 y + 17 = 0

x^{2} + 2 y^{2} - 3 x + 2 y - 2 = 0

What is ((x-2)^2)/9+((y-3)^2)/(25)=1 ?
The solution to ((x-2)^2)/9+((y-3)^2)/(25)=1 is Ellipse with (h,k)=(2,3),a=3,b=5