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Focal Radius

Focal Radius

This term has distinctly different definitions for different authors.

Usage 1: For some authors, this refers to the distance from the center to the focus for either an ellipse or a hyperbola. This definition of focal radius is usually written c.

Usage 2: For other authors, focal radius refers to the distance from a point on a conic section to a focus. In this case the focal radius varies depending where the point is on the curve (unless the conic in question is a circle). If there are two foci then there are two focal radii.

Note: Using this second definition, the sum of the focal radii of an ellipse is a constant. It is the same as the length of the major diameter. The difference of the focal radii of a hyperbola is a constant. It is the distance between the vertices.

 

Movie clip

Ellipse (pink) with a horizontal blue line (major axis) and a red vertical line segment showing a point on the curve with its...

Ellipse: Sum of the focal radii (usage 2) is constant
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See also

Foci of an ellipse, foci of a hyperbola, focus of a parabola

Key Formula

Ellipse: r1+r2=2aHyperbola: ∣r1−r2∣=2a\text{Ellipse: } r_1 + r_2 = 2a \qquad\qquad \text{Hyperbola: } |r_1 - r_2| = 2a
Where:
  • r1r_1 = Focal radius from a point on the curve to one focus
  • r2r_2 = Focal radius from the same point to the other focus
  • aa = Semi-major axis length (ellipse) or semi-transverse axis length (hyperbola)
  • 2a2a = Full length of the major axis (ellipse) or distance between vertices (hyperbola)

Worked Example

Problem: An ellipse has the equation x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1. A point P(3,125)P(3, \frac{12}{5}) lies on the ellipse. Find both focal radii of PP.
Step 1: Identify aa and bb from the equation. Here a2=25a^2 = 25 and b2=9b^2 = 9.
a=5,b=3a = 5, \quad b = 3
Step 2: Find cc, the distance from the center to each focus, using c2=a2−b2c^2 = a^2 - b^2.
c=25−9=16=4c = \sqrt{25 - 9} = \sqrt{16} = 4
Step 3: The foci are at (±4,0)(\pm 4, 0). Compute r1r_1, the distance from P(3,125)P(3, \frac{12}{5}) to F1(−4,0)F_1(-4, 0).
r1=(3−(−4))2+(125)2=49+14425=1225+14425=136925=375r_1 = \sqrt{(3-(-4))^2 + \left(\tfrac{12}{5}\right)^2} = \sqrt{49 + \tfrac{144}{25}} = \sqrt{\tfrac{1225 + 144}{25}} = \sqrt{\tfrac{1369}{25}} = \frac{37}{5}
Step 4: Compute r2r_2, the distance from PP to F2(4,0)F_2(4, 0).
r2=(3−4)2+(125)2=1+14425=16925=135r_2 = \sqrt{(3-4)^2 + \left(\tfrac{12}{5}\right)^2} = \sqrt{1 + \tfrac{144}{25}} = \sqrt{\tfrac{169}{25}} = \frac{13}{5}
Step 5: Verify: the sum of the focal radii should equal 2a=102a = 10.
r1+r2=375+135=505=10=2a  ✓r_1 + r_2 = \frac{37}{5} + \frac{13}{5} = \frac{50}{5} = 10 = 2a \; \checkmark
Answer: The two focal radii are r1=375=7.4r_1 = \frac{37}{5} = 7.4 and r2=135=2.6r_2 = \frac{13}{5} = 2.6. Their sum equals 2a=102a = 10, confirming the ellipse property.

Another Example

This example applies the focal radius concept to a hyperbola instead of an ellipse, showing that the key property changes from a constant sum to a constant absolute difference.

Problem: A hyperbola has the equation x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1. A point P(5,163)P(5, \frac{16}{3}) lies on the hyperbola. Find both focal radii and verify the constant-difference property.
Step 1: Identify aa and bb, then find cc.
a=3,b=4,c=9+16=5a = 3, \quad b = 4, \quad c = \sqrt{9 + 16} = 5
Step 2: The foci are at (±5,0)(\pm 5, 0). Compute r1r_1 from P(5,163)P(5, \frac{16}{3}) to F1(−5,0)F_1(-5, 0).
r1=(5+5)2+(163)2=100+2569=900+2569=11569=343r_1 = \sqrt{(5+5)^2 + \left(\tfrac{16}{3}\right)^2} = \sqrt{100 + \tfrac{256}{9}} = \sqrt{\tfrac{900 + 256}{9}} = \sqrt{\tfrac{1156}{9}} = \frac{34}{3}
Step 3: Compute r2r_2 from PP to F2(5,0)F_2(5, 0).
r2=(5−5)2+(163)2=2569=163r_2 = \sqrt{(5-5)^2 + \left(\tfrac{16}{3}\right)^2} = \sqrt{\tfrac{256}{9}} = \frac{16}{3}
Step 4: Verify: the absolute difference should equal 2a=62a = 6.
∣r1−r2∣=343−163=183=6=2a  ✓|r_1 - r_2| = \frac{34}{3} - \frac{16}{3} = \frac{18}{3} = 6 = 2a \; \checkmark
Answer: The focal radii are r1=343≈11.33r_1 = \frac{34}{3} \approx 11.33 and r2=163≈5.33r_2 = \frac{16}{3} \approx 5.33. Their difference equals 2a=62a = 6.

Frequently Asked Questions

What is the difference between a focal radius and the value cc in a conic section?
This depends on which definition of focal radius you are using. Under Usage 1, the focal radius is exactly cc, the fixed distance from the center to a focus. Under Usage 2, the focal radius is the variable distance from any point on the curve to a focus, which changes as the point moves. Most modern textbooks use Usage 2, so check which convention your course follows.
Why is the sum of the focal radii of an ellipse constant?
An ellipse is defined as the set of all points whose distances to two foci add up to a constant. That constant equals 2a2a, the length of the major axis. No matter where you pick a point on the ellipse, r1+r2=2ar_1 + r_2 = 2a always holds. This is the defining geometric property of an ellipse.
Does a parabola have focal radii?
A parabola has only one focus, so there is only one focal radius for any point on the curve. For a parabola, the focal radius equals the distance from the point to the directrix. This property is used directly in the reflective definition of a parabola: every point is equidistant from the focus and the directrix.

Focal radius (Usage 2) for an ellipse vs. Focal radius (Usage 2) for a hyperbola

Focal radius (Usage 2) for an ellipseFocal radius (Usage 2) for a hyperbola
DefinitionDistance from a point on the ellipse to a focusDistance from a point on the hyperbola to a focus
Key propertySum of two focal radii is constant: r1+r2=2ar_1 + r_2 = 2aAbsolute difference of two focal radii is constant: ∣r1−r2∣=2a|r_1 - r_2| = 2a
What 2a2a representsLength of the major axisDistance between the two vertices
Relationship c2=…c^2 = \ldotsc2=a2−b2c^2 = a^2 - b^2 (so c<ac < a)c2=a2+b2c^2 = a^2 + b^2 (so c>ac > a)

Why It Matters

Focal radii appear throughout precalculus and analytic geometry whenever you work with ellipses, hyperbolas, or parabolas. The constant-sum and constant-difference properties are essential for deriving the standard equations of these conics and for solving problems involving orbits, satellite dishes, and optics. Understanding focal radii also gives you a geometric way to verify whether a given point actually lies on a conic section.

Common Mistakes

Mistake: Using c2=a2−b2c^2 = a^2 - b^2 for a hyperbola instead of c2=a2+b2c^2 = a^2 + b^2.
Correction: For an ellipse, c2=a2−b2c^2 = a^2 - b^2 because the foci lie inside the curve. For a hyperbola, c2=a2+b2c^2 = a^2 + b^2 because the foci lie outside the vertices. Mixing these up gives the wrong focal distance and incorrect focal radii.
Mistake: Confusing the two usages: treating the fixed distance cc (center-to-focus) as if it were the variable focal radius from a point on the curve.
Correction: Always clarify which definition your textbook uses. Usage 1 (cc) is a single fixed number for a given conic. Usage 2 (r1r_1, r2r_2) varies depending on which point on the curve you choose.

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