What is ((x-1)^2)/(36)+((y+2)^2)/(16)=1 ?

The solution to ((x-1)^2)/(36)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(1,-2),a=6,b=4

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((x-1)^2)/(36)+((y+2)^2)/(16)=1

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{16} = 1

(h, k) = (1, - 2), a = 6, b = 4

Solution steps

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{6 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 6, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 6, b = 4

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

3 X^{2} + 2 Y^{2} = 8

9 x^{2} + 36 y^{2} = 324

5 x^{2} + 9 y^{2} - 20 x - 18 y - 16 = 0

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

What is ((x-1)^2)/(36)+((y+2)^2)/(16)=1 ?
The solution to ((x-1)^2)/(36)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(1,-2),a=6,b=4