What is 2(x+4)^2+(y+2)^2=8 ?

The solution to 2(x+4)^2+(y+2)^2=8 is Ellipse with (h,k)=(-4,-2),a=2,b=2sqrt(2)

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2(x+4)^2+(y+2)^2=8

2 (x + 4)^{2} + (y + 2)^{2} = 8

(h, k) = (- 4, - 2), a = 2, b = 22

Solution steps

2 (x + 4)^{2} + (y + 2)^{2} = 8

Rewrite

2 (x + 4)^{2} + (y + 2)^{2} = 8

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 4, - 2), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, - 2), b = 22, a = 2

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

\frac{x ^{2}}{2} + \frac{y ^{2}}{16} = 1

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

4 x^{2} + 24 y^{2} = 100

- x^{2} - 2 y^{2} = - 40

What is 2(x+4)^2+(y+2)^2=8 ?
The solution to 2(x+4)^2+(y+2)^2=8 is Ellipse with (h,k)=(-4,-2),a=2,b=2sqrt(2)