• Algebra > Finding the Center and Foci of an Ellipse

Finding the Center and Foci of an Ellipse 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.

How do we find the center of an ellipse given any random ellipse using a straight edge and a compass? Drawing tangents is not allowed since it may not be accurate.

If we can find the center, we can find the foci. Finding the center involves knowing that the midpoints of two sets of parallel chords of an ellipse are collinear with the center. The constuction is the easy part. Proving the midpoints of two parallel ellipse chords are collinear with the center is the hard part.

Figure 1 shows 2 sets of parallel chords of an ellipse with their midpoints. The slopes of these 2 sets of lines is different. The midpoints have been connected by a line. The intersection of the lines passes through the center of the ellipse. This is a property of ellipses which we will prove.

Once we find the center, we can use the fact that \( a^{2} = b^{2} + c^{2} \) to find the location of foci by drawing a right triangle with these lengths, which is not that difficult.

Proof

First, let’s state the given. The slope of the parallel chords AB and CD is equal since they are parallel. The slope is \( m = \frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \frac{y_{3}-y_{4}}{x_{3}-x_{4}} \). Keep this in our data bank.

Second, the midpoint of AB is given by \( m_{1,2} = \left( \frac{x_{1}+x_{2}}{2}, \, \frac{y_{1}+y_{2}}{2} \right) \). Let’s also keep this in mind.

Next, we know that both A and B are on the ellipse, so they have to satisfy the equations \( \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} = 1 \) and \( \frac{x_{2}^{2}}{a^{2}} + \frac{y_{2}^{2}}{b^{2}} = 1 \). Let’s subtract these two and manipulate them as follows.

(i) \( \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - \frac{x_{2}^{2}}{a^{2}} - \frac{y_{2}^{2}}{b^{2}} = 1 - 1 \)

(ii) \( \frac{x_{1}^{2} - x_{2}^{2}}{a^{2}} + \frac{y_{1}^{2} - y_{2}^{2}}{b^{2}} = 0 \)

(iii) \( b^{2}(x_{1}^{2} - x_{2}^{2}) + a^{2}(y_{1}^{2} - y_{2}^{2}) = 0 \)  (Multiplying by \(a^{2}b^{2} \))

(iv) \( b^{2}(x_{1} - x_{2})(x_{1} + x_{2}) + a^{2}(y_{1} - y_{2})(y_{1} + y_{2}) = 0 \)

Remember the midpoint m_1,2 we found above was \( m_{1,2} = \left( \frac{x_{1}+x_{2}}{2}, \, \frac{y_{1}+y_{2}}{2} \right) \). Let’s call \( x_{m} \) the x value of the midpoint and \( y_{m} \) the y value of the midpoint. Then \( 2x_{m} = x_{1} + x_{2} \) and \( 2y_{m} = y_{1} + y_{2} \). Substituting these in (iv) gives us:

(v) \( 2x_{m}b^{2}(x_{1} - x_{2}) + 2y_{m}a^{2}(y_{1} - y_{2}) = 0 \)

(vi) \( x_{m}b^{2}(x_{1} - x_{2}) + y_{m}a^{2}(y_{1} - y_{2}) = 0 \)  (Dividing by 2)

Now, recall the slope of the chords was \( m = \frac{y_{1}-y_{2}}{x_{1}-x_{2}} \). That means \( m({x_{1}-x_{2}}) = y_{1}-y_{2} \). Making that substitution in (vi) gives us:

(vii) \( x_{m}b^{2}(x_{1} - x_{2}) + y_{m}a^{2}m({x_{1}-x_{2}}) = 0 \)

(viii) \( x_{m}b^{2} + y_{m}a^{2}m = 0 \)

(ix) \( y_{m} = -\frac{b^{2}}{a^{2}}\cdot \frac{1}{m}\cdot x_{m} \)

Equation (ix) represents the relationship between the midpoint of the chord with a, b and the slope of the chord. We can determine either the x value or y if we know the other. We can use this relationship for any arbitrary chord when we know the slope.

Now, let’s consider the midpoints m_1,2 and m_3,4 of AB and CD, respectively, using the above formula we got. We can determine the slope of the line that passes through the midpoints by using the above relationships. The slope, \( m_{L} \), of the line will be \( m_{L} = \frac{y_{m1} - y_{m2}}{x_{m1} - x_{m2}} \) where m₁ and m₂ are the two midpoints respectively. Using (ix), we will substitute to find the slope of the line through the midpoints.

\( m_{L} = \frac{y_{m1} - y_{m2}}{x_{m1} - x_{m2}} \)

(x) \( m_{L} = \dfrac{-\frac{b^{2}}{a^{2}}\cdot \frac{1}{m}\cdot \frac{x_{3}-x_{4}}{2} + \frac{b^{2}}{a^{2}}\cdot \frac{1}{m}\cdot \frac{x_{1}-x_{2}}{2}}{\frac{x_{3}-x_{4}}{2} - \frac{x_{1}-x_{2}}{2}} \)

(xi) \( m_{L} = -\dfrac{b^{2}}{a^{2}}\cdot \frac{1}{m} \cdot \frac{\frac{x_{3}-x_{4}}{2} -\cdot \frac{x_{1}-x_{2}}{2}}{\frac{x_{3}-x_{4}}{2} - \frac{x_{1}-x_{2}}{2}} \)

(xii) \( m_{L} = -\frac{b^{2}}{a^{2}}\cdot \frac{1}{m} \)

We know the slope of the line that passes through the midpoints and we know 2 points on the line. We only need to know 1 point on the line, but we can use any arbitrary midpoint \( (x_{a}, y_{a}) \) to find the equation of the line. The arbitrary midpoint has to satisfy \( y_{m} = -\frac{b^{2}}{a^{2}}\cdot \frac{1}{m}\cdot x_{m} \).

The equation of the line \( y - y_{a} = -\frac{b^{2}}{ma^{2}}(x - x_{a}) \) becomes \( y + \frac{b^{2}}{ma^{2}}x_{a} = -\frac{b^{2}}{ma^{2}}(x - x_{a}) \).

(xiii) \( y + \frac{b^{2}}{ma^{2}}x_{a} = -\frac{b^{2}}{ma^{2}}(x - x_{a}) \)

(xiv) \( y + \frac{b^{2}}{ma^{2}}x_{a} = -\frac{b^{2}}{ma^{2}}x + \frac{b^{2}}{ma^{2}}x_{a} \)

(xv) \( y = -\frac{b^{2}}{ma^{2}}x \)

This actually proves that the line that passes through the midpoint of parallel chords passes through the center since the equation of the line has no y-intercept. The point (0, 0) is a point on this line.

Now that we know how to find the center and proved it, we can find the vertices and the foci.

The Vertices

To find the foci, we need to find the vertices first. We cannot simply eyeball it; rather, we need to use a straight edge and compass to get it as precise as possibe.

Step 1

The first step is to draw a circle at the center we just found. Let the diameter be greater than the minor axis. Draw a chord through two of the four points of intersection of the circle and the ellipse.

Step 2

Next, draw a perpendicular to the chord RS at the midpoint of RS. This line with intersect the ellipse at the major vertices.

I also constructed a perpendicular to the major axis through the center, which will intersect at the minor vertices. So, now we have all 4 vertices.

Step 3

In the last step, take the measure of TG, which is the length of half the major axis, or a, and draw an arc from either minor vertex onto the major axis. The intersection is labeled F₁.

Since VG = b and VF₁ = a, GF₁ has to equal c because of the right triangle equality \( a^{2} = b^{2} + c^{2} \) as shown in Figure 5. In the same way, we can also location F₂.

Thus completes our construction of finding the center, the four vertices, and the foci of an ellipse.

The reader is left to draw the two directrices of the ellipse as a challenge.

Anirdesh.com Last Updated: May 27 2025, 01:47:36.