What is ((x-5)^2)/4+((y+2)^2)/2 =1 ?

The solution to ((x-5)^2)/4+((y+2)^2)/2 =1 is Ellipse with (h,k)=(5,-2),a=2,b=sqrt(2)

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((x-5)^2)/4+((y+2)^2)/2 =1

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

(h, k) = (5, - 2), a = 2, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 2), a = 2, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 2), a = 2, b = 2

6 x^{2} + 144 y^{2} = 864

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

81 x^{2} + 9 y^{2} = 64

y^{2} + 30 x^{2} = 30

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

What is ((x-5)^2)/4+((y+2)^2)/2 =1 ?
The solution to ((x-5)^2)/4+((y+2)^2)/2 =1 is Ellipse with (h,k)=(5,-2),a=2,b=sqrt(2)