What is the axis y^2-4y-8x-12=0 ?

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axis y^2-4y-8x-12=0

axis

y^{2} - 4 y - 8 x - 12 = 0

y = 2

Solution steps

y^{2} - 4 y - 8 x - 12 = 0

Rewrite

y^{2} - 4 y - 8 x - 12 = 0

in the standard form

4 \cdot 2 (x - (- 2)) = (y - 2)^{2}

Therefore parabola properties are:

(h, k) = (- 2, 2), p = 2

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 2

y^{2} = 16 x - 8

y = x^{2} + 10 x - 11

x = (y + 4)^{2}

y = + x^{2} - 9

y = + \frac{9 x ^{2} - 6 x + 3}{3} x