What is ((x+1)^2)/(36)+((y+3)^2)/(25)=1 ?

The solution to ((x+1)^2)/(36)+((y+3)^2)/(25)=1 is Ellipse with (h,k)=(-1,-3),a=6,b=5

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((x+1)^2)/(36)+((y+3)^2)/(25)=1

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 3 ) ^{2}}{25} = 1

(h, k) = (- 1, - 3), a = 6, b = 5

Solution steps

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 3 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{6 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, - 3), a = 6, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 3), a = 6, b = 5

x^{2} + 9 y^{2} + 2 x - 18 y + 1 = 0

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{80} = 1

4 x^{2} + y^{2} - 16 x + 2 y + 1 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{36} = 1

What is ((x+1)^2)/(36)+((y+3)^2)/(25)=1 ?
The solution to ((x+1)^2)/(36)+((y+3)^2)/(25)=1 is Ellipse with (h,k)=(-1,-3),a=6,b=5