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

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

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

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

(h, k) = (3, 1), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 1), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 1), a = 5, b = 3

9 x^{2} - 18 x + 4 y^{2} + 24 y = - 9

\frac{x}{x ^{2} + y ^{2}} = 2

\frac{x ^{2}}{200} + \frac{y ^{2}}{50} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{144} = 1

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

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