What is 2x^2+y^2+16x-4y+32=0 ?

The solution to 2x^2+y^2+16x-4y+32=0 is Ellipse with (h,k)=(-4,2),a=sqrt(2),b=2

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2x^2+y^2+16x-4y+32=0

2 x^{2} + y^{2} + 16 x - 4 y + 32 = 0

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

Solution steps

2 x^{2} + y^{2} + 16 x - 4 y + 32 = 0

Rewrite

2 x^{2} + y^{2} + 16 x - 4 y + 32 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x - 5 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{4} = 1

4 y^{2} - 338 x + 32 y = - 169 x^{2} + 443

2 x^{2} + y^{2} - 4 x + 8 y - 4 = 0

25 x^{2} + 9 y^{2} - 250 x - 18 y + 409 = 0

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

What is 2x^2+y^2+16x-4y+32=0 ?
The solution to 2x^2+y^2+16x-4y+32=0 is Ellipse with (h,k)=(-4,2),a=sqrt(2),b=2