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

The solution to (x^2)/(36)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(0,-1),a=6,b=5

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

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

(h, k) = (0, - 1), a = 6, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{6 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 6, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), a = 6, b = 5

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

25 x^{2} + 16 y^{2} + 100 x + 32 y - 284 = 0

4 X^{2} + 8 Y^{2} - 24 X + 32 Y - 12 = 0

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{20} = 1

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

What is (x^2)/(36)+((y+1)^2)/(25)=1 ?
The solution to (x^2)/(36)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(0,-1),a=6,b=5