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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

y + 2 = \frac{6 - ( \frac{6 y + 2 x}{2 + y} )}{2} x

42 x^{2} + 49 y^{2} = 4

\frac{x ^{2}}{C 1 ^{2}} = 1 - \frac{y ^{2}}{C 2 ^{2}}

\frac{( x - 1 ) ^{2}}{169} + \frac{( y - 4 ) ^{2}}{144} = 1

9 x^{2} + 25 y^{2} - 18 x - 50 y - 191 = 0

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