What is 4(x-3)^2+2y^2=8 ?

The solution to 4(x-3)^2+2y^2=8 is Ellipse with (h,k)=(3,0),a=sqrt(2),b=2

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4(x-3)^2+2y^2=8

4 (x - 3)^{2} + 2 y^{2} = 8

(h, k) = (3, 0), a = 2, b = 2

Solution steps

4 (x - 3)^{2} + 2 y^{2} = 8

Rewrite

4 (x - 3)^{2} + 2 y^{2} = 8

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{( 2 ) ^{2}} + \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 0), a = 2, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 0), b = 2, a = 2

9 x^{2} + 36 y^{2} + 18 x - 360 y + 585 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{24} = 1

x^{(2)} + 4 y^{(2)} - 6 x + 16 y + 21 = 0

16 x^{2} + 25 y^{2} - 64 x - 336 = 0

x^{2} + 2 y^{2} - 4 y = 6

What is 4(x-3)^2+2y^2=8 ?
The solution to 4(x-3)^2+2y^2=8 is Ellipse with (h,k)=(3,0),a=sqrt(2),b=2