Parabola with Shifted Vertex

Vertex at (h, k), axis parallel to x-axis. All properties of y² = 4ax apply with origin shifted to (h, k). Similarly (x−h)² = 4a(y−k) for axis parallel to y-axis.

When the vertex is at $(h, k)$ and the axis is parallel to the x-axis, substitute $X = x - h$, $Y = y - k$ to translate the origin to $(h, k)$. In the new coordinates the equation is $Y^2 = 4aX$, which translates back to:

$$(y - k)^2 = 4a(x - h)$$

Elements in original coordinates:

Similarly, $(x - h)^2 = 4a(y - k)$ has vertex $(h, k)$, focus $(h, k+a)$, directrix $y = k - a$.

Identifying the vertex from a general equation: Any equation of the form $y^2 + Dy + Ex + F = 0$ (with no $x^2$ term) represents a parabola. Complete the square in $y$:

$$(y + D/2)^2 = -Ex - F + D^2/4$$ $$(y + D/2)^2 = -E\!\left(x - \frac{D^2/4 - F}{E}\right)$$

This is $(Y)^2 = -E \cdot X$, a parabola with vertex at $\left(\frac{D^2 - 4F}{4E},\, -\frac{D}{2}\right)$.