What is 2x-x^2-4y^2-16y-1>= 0 ?

The solution to 2x-x^2-4y^2-16y-1>= 0 is Ellipse with (h,k)=(1,-2),a=4,b=2

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2x-x^2-4y^2-16y-1>= 0

2 x - x^{2} - 4 y^{2} - 16 y - 1 \geq 0

(h, k) = (1, - 2), a = 4, b = 2

Solution steps

2 x - x^{2} - 4 y^{2} - 16 y - 1 \geq 0

Rewrite

2 x - x^{2} - 4 y^{2} - 16 y - 1 \geq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 4, b = 2

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

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

6 x^{2} + y^{2} + 10 x = 0

x^{2} + 4 y^{2} - 4 \geq 0

2 x^{2} + 3 y^{2} - 6 x + 4 - y = 0

What is 2x-x^2-4y^2-16y-1>= 0 ?
The solution to 2x-x^2-4y^2-16y-1>= 0 is Ellipse with (h,k)=(1,-2),a=4,b=2