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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

25 (x^{2} + 4 x + 4) + 36 (y^{2} + 2 y + 1) = 770

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

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

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