What is the axis y^2-10x+4y-16=0 ?

The axis y^2-10x+4y-16=0 is y=-2

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axis y^2-10x+4y-16=0

axis

y^{2} - 10 x + 4 y - 16 = 0

y = - 2

Solution steps

y^{2} - 10 x + 4 y - 16 = 0

Rewrite

y^{2} - 10 x + 4 y - 16 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 2, - 2), p = \frac{5}{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 = - 12 x^{2} + 156 x

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

y = 2 x^{2} - x

(x + 3)^{2} = 4 (y - 1)

y = x^{2} + 2 x + 1