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

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

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

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

(h, k) = (1, - 2), a = 36, b = 25

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 36, b = 25

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 36, b = 25

9 x^{2} + 36 y^{2} + 36 x - 216 y + 36 = 0

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

25 x^{2} + 16 y^{2} - 400 = 0

4 x^{2} + 9 y^{2} - 24 x - 36 y + 36 = 0

x^{2} + 9 y^{2} = 3

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