What is the axis y^2+6x-4y-11=0 ?

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axis y^2+6x-4y-11=0

axis

y^{2} + 6 x - 4 y - 11 = 0

y = 2

Solution steps

y^{2} + 6 x - 4 y - 11 = 0

Rewrite

y^{2} + 6 x - 4 y - 11 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (\frac{5}{2}, 2), p = - \frac{3}{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

(x + 3)^{2} = 2 (y - 4)

+ 3 y^{2} = 24 x

y = - 2 x^{2} - 8 x - 3

y = - \frac{13}{11} (x + 3)^{2} + 1

y = (x - 3)^{2} - 2