What is the axis y^2+x-10y+29=0 ?

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axis y^2+x-10y+29=0

axis

y^{2} + x - 10 y + 29 = 0

y = 5

Solution steps

y^{2} + x - 10 y + 29 = 0

Rewrite

y^{2} + x - 10 y + 29 = 0

in the standard form

4 (- \frac{1}{4}) (x - (- 4)) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (- 4, 5), p = - \frac{1}{4}

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 = 5

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

y = 9 - (x - 2)^{2}

y^{2} + 8 x + 2 y + 33 = 0

y = 3 x^{2} - 6 x - 2

y = 3 x^{2} - 6 x - 1