Finding the Vertex and Focus of a Parabola by Construction

We can find the center of a circle by drawing two chords of a circle, then constructing the perpendiculars of the chords. The intersection will be the center. We can also find the foci and vertices of any arbitrary ellipse by construction by drawing 2 sets of parallel chords and connecting their midpoints.

This page will show the steps to finding the vertex and focus of a parabola and the proof that the midpoints of two chords of a parabola pass through a line that is parallel to the vertex.

Figure 1: Parallel chords of a parabola

Figure 1 shows two parallel chords of a parabola. The midpoints are connected by a line. This line is parallel to the line of symmetry of the parabola. Hence, we can draw a perpendicular chord to this line. The midpoint of this chord will be collinear to the vertex.

Figure 2: Parallel chords of a parabola

We will prove this first before proceeding to find the vertex and the focus.

PROOF

In Figure 2, points A and B are on a chord of the parabola. We will find the midpoint first. The midpoint is \( \left( \frac{x_{1}+x_{2}}{2},\, \frac{y_{1}+y_{2}}{2} \right) \). Since the points are on a parabola, let’s define the y values as being on a parabola that has the equation \( y = ax^{2} \). Therefore, the midpoint is \( \left( \frac{x_{1}+x_{2}}{2},\, \frac{ax_{1}^{2}+ax_{2}^{2}}{2} \right) \).

The slope, m, of the chord AB is given by \( m = \frac{ax_{2}^{2} - ax_{1}^{2}}{x_{2} - x_{1}} \). This can be factored as \( m = a\frac{(x_{2}+x_{1})(x_{2}-x_{1})}{x_{2} - x_{1}} = a(x_{1}+x_{2}) \).

Let \( x_{m} \) be the x value of the midpoint of chord AB. Then, \( x_{1} + x_{2} = 2ax_{m} \). Therefore, the slope of chord AB is \( m = 2ax_{m} \).

The slope, m, of a chord of a parabola \( y = ax^{2} \) is given by \( m = 2ax_{m} \), where \( x_{m} \) is the x value of the midpoint of the chord.

What this means is that if we choose a slope, then the midpoint is predetermined, because \( x_{m} = \frac{m}{2a} \) also holds true. Any chord with a slope of m will have their midpoint at the same x value. For a parabola defined as \( y = ax^{2} \), then this represents the line \( x = \frac{m}{2a} \), which is parallel to the y-axis and the line of symmetry. Even if the parabola is rotated, these properties have to remain intact, meaning that the line through the midpoints remain parallel to the line of symmetry. Therefore, this actually completes the proof.

THE FOCUS

The vertex was quite trivial. We found it above from the intersection of the line of symmetry and the parabola. Now, we will find the focus. Some methods will instruct you to construct a tangent at the perpendicular chord. However, since we already located the vertex, we can simply draw a perpendicular there. This line will represent the x-axis. The line of symmetry will represent the y-axis.

Figure 3: The Axes

Next, pick any random point on the x-axis and draw a perpendicular there. It will fall on the parabola at point P. Then find the midpoint, M, of the segment OR. We are going to connect M and P, which will be the tangent at P because of the properties of parabolas: A tangent to the parabola crosses the x-axis at the midpoint of the x value of the point of tangency.

Figure 4: Y-Intercept Construction

Construct a line that passes throught P and M. This line will be tangent to the parabola at P and will cross the x-axis at M which is the midpoint of OR.

Figure 5: A Parabola Tangent

Lastly, draw a perpendicular to the tangent at M on the x-intercept. This line will intersect the line of symmetry at the focus point. As you can see in the image, the lengths PN and FP are equal, which defines a parabola. A line through N parallel to the x-axis is the directrix of the parabola.

Figure 6: The Focus and the Directrix

This completes how to find the vertex and the focus of a parabola. We used all valid methods of geometrical construction using a straight edge and a compass.

Here is a Geogebra activity that let’s you play with the points on a parabola to show the line connecting the midpoints of the chords remains parallel to the line of symmetry.

You can drag points A, B, and C. Point D remains fixed to maintain parallel chords.