Focus of a Parabola

By Daniel Machado Updated: 11/02/2026

The focus of a parabola is a fixed point located on the axis of symmetry around which the curve wraps. This point appears in the definition of a parabola, as this conic section is the set of all points that are equidistant from the focus and a fixed line called the directrix.

The vertex is located at the same distance from the focus as it is from the directrix. This distance is |p|, where "p" is the parameter that appears in the standard equation. We use this information to find the coordinates of the focus:

For a horizontal parabola with vertex (h, k) and equation (y - k)² = 4p (x - h), the focus is at the coordinates (h + p, k).
For a vertical parabola with vertex at (h, k) and equation (x - h)²= 4p (y - k), the focus is located at (h, k + p).

If the vertex is at the origin, the focus is at (p, 0) for horizontal parabolas and (0, p) for vertical parabolas.

Note: Here we focus only on one element of the parabola. To gain a complete understanding, we recommend reviewing our main guide on this conic section first, where we cover its fundamental theory:

How to Find the Focus

As shown above, finding the focus requires determining the value of parameter p along with its sign. The standard formulas are summarized in the table below.

Parabola Equation	Vertex	Orientation	Focus
y² = 4px	(0, 0)	Horizontal	(p, 0)
x²= 4py	(0, 0)	Vertical	(0, p)
(y - k)² = 4p (x - h)	(h, k)	Horizontal	(h + p, k)
(x - h)²= 4p (y - k)	(h, k)	Vertical	(h, k + p)

When given the general form of the equation, it is necessary to complete the square to convert it into standard form and find the value of parameter p.

Examples

Find the coordinates of the focus for each of the following parabolas:

y² = 8x
x² = -2y
(x + 2)² = 6(y - 3)
(y - 2)² = 4(x - 1)
x² - 8x + 10y + 66 = 0
y² + 6x - 8y + 4 = 0

Solution 1

Given equation: \(y^2=8x\)

First, we identify that the parabola is horizontal and has its vertex at the origin (0, 0). The standard form of a horizontal parabola is \(y^2=4px.\) Comparing this with the given equation, \(4p=8,\) we find that \(p=2.\) Since the focus of a horizontal parabola with vertex at the origin is located at (p, 0), the coordinates of the focus are (2, 0).

Graph of the focus of a parabola, Example 1

Solution 2

Given equation: \(x^2=-2y\)

This is a vertical parabola with its vertex at (0, 0). The standard form is \(x^2=4py.\) Setting \(4p=-2,\) we calculate \(p=-0.5.\) For a vertical parabola with vertex at the origin, the focus is located at \((0, p).\) Therefore, the focus is at \((0, -0.5).\)

Graph of the focus of a parabola, Example 2

Solution 3

Given equation: \((x+2)^2=6 (y-3)\)

This is a vertical parabola with its vertex at (-2, 3). The standard form is \((x-h)^2=4p (y-k).\) Comparing terms, \(4p=6,\) which gives \(p=1.5.\) Since the focus of a vertical parabola is at \((h, k+p),\) we find the coordinates to be \((-2, 3+1.5)=(-2, 4.5).\)

Graph of the focus of a parabola, Example 3

Resource: If you are solving practice problems and want to verify your answers, you can use our free tool:

Solution 4

Given equation: \((y-2)^2=4 (x-1)\)

This is a horizontal parabola with vertex at (1, 2). The standard form is \((y-k)^2=4p (x-h).\) Here, \(4p=4,\) so \(p=1.\) For a horizontal parabola, the focus is at \((h+p, k).\) Thus, the focus is located at \((1+1, 2)=(2, 2).\)

Graph of the focus of a parabola, Example 4

Solution 5

Given equation: \(x^2-8x+10y+66=0\)

The equation is in general form, but by completing the square we obtain the standard form: \((x-4)^2=-10 (y+5).\)

This parabola is vertical with vertex at (4, -5). The standard form is \((x-h)^2=4p (y-k).\) Comparing, \(4p=-10,\) yielding \(p=-2.5.\) Because the focus of a vertical parabola is located at \((h, k+p),\) the coordinates are \((4, -5-2.5)=(4, -7.5).\)

Graph of the focus of a parabola, Example 5

Solution 6

Given equation: \(y^2+6x-8y+4=0\)

Completing the square gives the standard form: \((y-4)^2=-6 (x-2).\)

We identify a horizontal parabola with vertex at (2, 4), whose standard form is \((y-k)^2=4p (x-h).\) Comparing, \(4p=-6,\) so \(p=-1.5.\) Since the focus of a horizontal parabola is at \((h+p, k),\) substituting these values yields \((2-1.5, 4)=(0.5, 4).\)