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

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

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

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

(h, k) = (2, - 1), a = 3, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 3, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), a = 3, b = 2

\frac{( x - 4 ) ^{2}}{12} + \frac{( y + 3 ) ^{2}}{16} = 1

x^{2} + y^{2} = \frac{2 x}{x ^{2} + y ^{2}}

2 x^{2} - 4 x + y^{2} + 2 y - 5 = 0

4 x^{2} - 40 x + 9 y^{2} + 54 y + 145 = 0

9 X^{2} + Y^{2} = 81

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