Conics - Parabola: Vertex, Focus, and Directrix

In earlier algebra work, a parabola probably showed up as the graph of a quadratic function, \(y = ax^2 + bx + c\). In conic sections, a parabola gets a geometric definition instead: it is the set of all points that are the same distance from a fixed point (the focus) as they are from a fixed line (the directrix). Every point on the curve balances these two distances exactly, and that balance is what shapes the familiar U-curve.

This definition connects directly to two of its sibling conics: a circle is the set of points a fixed distance from one center point, while a parabola trades that single center for a point and a line. If you have not yet seen how the other conics compare, it is worth glancing at Conics - Ellipse for contrast before continuing here.

When the vertex is at the origin \((0,0)\), a parabola takes one of two basic standard forms, depending on whether it opens vertically or horizontally. The number \(p\) is the distance from the vertex to the focus (and also from the vertex to the directrix).

Opens	Standard equation	Focus	Directrix
Up (\(p>0\)) or down (\(p<0\))	\(x^2 = 4py\)	\((0, p)\)	\(y = -p\)
Right (\(p>0\)) or left (\(p<0\))	\(y^2 = 4px\)	\((p, 0)\)	\(x = -p\)

When the vertex is shifted to a point \((h,k)\) instead of the origin, replace \(x\) with \((x-h)\) and \(y\) with \((y-k)\):

\((x-h)^2 = 4p(y-k)\) for vertical parabolas, or \((y-k)^2 = 4p(x-h)\) for horizontal ones.

Four features describe a parabola completely once you know its equation:

• Vertex: the turning point of the curve, at \((h,k)\).
• Axis of symmetry: the line through the vertex and focus that splits the parabola into two mirror halves. For a vertical parabola it is \(x = h\); for a horizontal one it is \(y = k\).
• Focus: the fixed point inside the curve used in the distance definition, located a distance \(|p|\) from the vertex along the axis of symmetry.
• Directrix: the fixed line outside the curve, also a distance \(|p|\) from the vertex, on the opposite side from the focus.

A related segment worth knowing is the latus rectum, the chord that passes through the focus perpendicular to the axis of symmetry. Its length is always \(|4p|\), which makes it a quick way to sketch how "wide" the parabola is right at the focus.

Consider \(x^2 = 8y\). Comparing this to \(x^2 = 4py\) gives \(4p = 8\), so \(p = 2\). The vertex is at the origin, the parabola opens upward (since \(p>0\)), the focus is at \((0,2)\), and the directrix is the horizontal line \(y = -2\). Solving for \(y\) gives \(y = \dfrac{x^2}{8}\), which is graphed below.

Find the vertex, focus, and directrix of \((x-2)^2 = 4(y+1)\).

Matching \((x-h)^2 = 4p(y-k)\), the vertex is \((h,k) = (2,-1)\). Since \(4p = 4\), \(p = 1\). Because \(p\) is positive and the squared term is in \(x\), the parabola opens upward. The focus is \(p = 1\) unit above the vertex, at \((2, 0)\), and the directrix is \(1\) unit below the vertex, the horizontal line \(y = -2\). The axis of symmetry is \(x = 2\).

Every quadratic function you have graphed as \(y = a(x-h)^2 + k\) is a parabola in disguise, just written to isolate \(y\) instead of keeping the squared term on one side. If you rearrange that form to match \((x-h)^2 = \dfrac{1}{a}(y-k)\), you can read off \(4p = \dfrac{1}{a}\) and find the focus and directrix directly. If you need a refresher on getting to that vertex form, or on turning general form into vertex form by completing the square, review those lessons first, since the conic-form vertex and focus formulas here build directly on them.

Once you are comfortable with parabolas, the same distance-based reasoning extends naturally to the other conic sections, including the hyperbola, which uses two foci instead of one.

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