What is the axis y^2+8x-4y+20=0 ?

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axis y^2+8x-4y+20=0

axis

y^{2} + 8 x - 4 y + 20 = 0

y = 2

Solution steps

y^{2} + 8 x - 4 y + 20 = 0

Rewrite

y^{2} + 8 x - 4 y + 20 = 0

in the standard form

4 (- 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 = x^{2} - 10 x + 12

y = 3 x^{2} + 6

y^{2} - 6 y - 4 x + 5 = 0

y = 5 x^{2} + x - 3

y = 7 x^{2} + 14 x