What is 4x^2+y^2+8x>0 ?

The solution to 4x^2+y^2+8x>0 is Ellipse with (h,k)=(-1,0),a=1,b=2

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4x^2+y^2+8x>0

4 x^{2} + y^{2} + 8 x > 0

(h, k) = (- 1, 0), a = 1, b = 2

Solution steps

4 x^{2} + y^{2} + 8 x > 0

Rewrite

4 x^{2} + y^{2} + 8 x > 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), b = 2, a = 1

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

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

200 = \frac{1000}{1 + x ^{2} + y ^{2}}

\frac{( x - 8 ) ^{2}}{42} + \frac{( y + 7 ) ^{2}}{16} = 1

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

What is 4x^2+y^2+8x>0 ?
The solution to 4x^2+y^2+8x>0 is Ellipse with (h,k)=(-1,0),a=1,b=2