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

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

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

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

(h, k) = (1, - 2), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 5, b = 3

49 x^{2} + 64 y^{2} = 25

- 2 x^{2} - 8 x - y^{2} + 16 y = 52

9 x^{2} + 4 y^{2} - 90 x + 32 y + 145 = 0

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

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

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