What is ((x+3)^2)/9+((y-1)^2)/(25)-1=0 ?

The solution to ((x+3)^2)/9+((y-1)^2)/(25)-1=0 is Ellipse with (h,k)=(-3,1),a=3,b=5

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((x+3)^2)/9+((y-1)^2)/(25)-1=0

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} - 1 = 0

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

Solution steps

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} - 1 = 0

Rewrite

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} - 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{4 - 16} = 1

\frac{x ^{2}}{2} + \frac{y ^{2}}{a - 16} = 1

x^{2} + 2 y^{2} - 8 y + 4 = 0

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

What is ((x+3)^2)/9+((y-1)^2)/(25)-1=0 ?
The solution to ((x+3)^2)/9+((y-1)^2)/(25)-1=0 is Ellipse with (h,k)=(-3,1),a=3,b=5