Derive both branches of a hyperbola

Try

Use foci F₁=(−5,0) and F₂=(5,0). This time require the absolute difference of distances to be 6. At P=(3,0), the distances are 8 and 2, so the difference is 6. At (−3,0) the distances exchange and the signed difference becomes −6, but its absolute value is still 6.

Both points belong. A definition using only r₁−r₂=6 would keep one branch and discard the other. The full focal hyperbola uses ∣r₁−r₂∣=6 and will lead to x²/9−y²/16=1.

Predict

The reverse triangle inequality bounds the absolute distance difference by the focal separation, here 10. Predict what happens if the required difference exceeds 10: no point can satisfy it. Equality gives rays on the focal line outside the segment between the foci, not a nondegenerate hyperbola.

In general take foci (−c,0),(c,0) and required difference 2a, with 0 < a < c. Define b²=c²−a²>0. The vertices will be (±a,0), while the foci lie farther from the center at (±c,0). Notice the contrast with an ellipse, where the focal distance c is smaller than the horizontal semiaxis a.

Experiment

Choose a and c in the explorer. Before revealing the distances, predict the sign of r₁−r₂ for a probe on the right branch. Reflect the probe across the vertical axis and predict what changes and what remains unchanged. Compare the signed and absolute differences explicitly.

Test a hyperbola against its focal definition

F1=(−c,0) and F2=(c,0). The positive target is 2a. A hyperbola requires |PF1−PF2|=2a, including both branches. Choose a point P and predict whether it meets that condition. The equation check is a separate representation, not a replacement for the focal definition.

Positive parameter a2345Focus distance c123456Reset focal explorer

c>a>0, so the hyperbola is nondegenerate and b²=c²−a². Here b²=16.

Probe xProbe yPlace P at right vertexPlace P at a nonvertex sampleReflect P across y-axisReveal focal checks

The solid curve is the focal locus, drawn with bounded samples and clipped to this fixed window. Hyperbola branches and degenerate rays continue beyond it. Dashed PF1 and dotted PF2 appear after revealing; the square and open-circle focus markers supplement color.

Predict the two distances and decide which combination must stay fixed. Reveal the calculations after committing to a prediction.

Use a nonvertex sample, reflect it across the y-axis and compare the signed and absolute distance differences. Change a and c to an equality case and an impossible case. Explain those cases using the triangle inequality before relying on a graph. Positive c keeps the two foci distinct in this explorer.

Coordinates and distances use floating-point arithmetic. Each control change hides old checks; this temporary exploration does not save a response or award mastery. Enter your derivation and its reverse implication in the written practice.

Test a vertex and a nonvertex point. Then try a=c and a>c, explaining the degenerate and empty outcomes from the distance inequality. The finite viewing window can show only part of an unbounded branch; an edge of the screen is not an endpoint of the curve.

Fixed example point	Distance to (−5,0)	Distance to (5,0)	Signed difference
(3,0)	8	2	6
(−3,0)	2	8	−6
(5,16/3)	34/3	16/3	6

Understand

Let r₁=√((x+c)²+y²) and r₂=√((x−c)²+y²). Their squared difference is 4cx, so the sign of r₁−r₂ is the sign of x: the distance sum is positive and c>0. Begin with the right branch, where r₁−r₂=2a. Factoring the squared difference gives r₁+r₂=2cx/a. Solving these two equations yields

r₁=cx/a+a, and r₂=cx/a−a.

Substitute the first into its squared distance formula. After canceling 2cx, obtain (c²−a²)x²/a²−y²=c²−a². Divide by b²=c²−a² to get x²/a²−y²/b²=1. Reflection across the vertical axis swaps the focal distances, so the left branch satisfies the same equation.

Now prove the reverse implication. Any point of x²/a²−y²/b²=1 satisfies ∣x∣≥a. For x≥a, the expressions cx/a+a and cx/a−a are positive: the smaller is at least c−a>0. Replacing y² using the equation shows

(x+c)²+y²=(cx/a+a)² and (x−c)²+y²=(cx/a−a)².

Because the unsquared expressions are positive, the actual distances are exactly those expressions. Their difference is 2a. For x≤−a, reflect the point to the right branch; the two distances swap, so the signed difference is −2a and its absolute value is 2a. This establishes the full two-branch equivalence without accepting arbitrary squared roots.

For a=3,c=5, b²=25−9=16. At the nonvertex point (5,16/3), the equation gives 25/9−16/9=1 and the table confirms the focal condition. That finite check illustrates the proof but does not replace it.

Apply

Given fresh foci and a positive distance difference, first check 0 < a < c. Derive the equation, identify both branches and vertices, and verify a nonvertex point in the original distance condition. A correct quadratic with an incorrect branch interpretation is an incomplete answer.

Diagnose the claim that the foci are the vertices. In the worked example the vertices are at ±3 and the foci at ±5. Also diagnose the use of b²=a²−c²: that ellipse relationship has the wrong sign here. The distance definition determines the parameter relationship, not a memorized visual resemblance.

Continue

An ellipse has a fixed sum and bounded locus; a hyperbola has a fixed absolute difference and unbounded branches. Both derivations require a reverse argument because squaring can lose sign information. The existing parabola lesson supplies the third distance-based conic, using a focus and a directrix rather than two foci.

Optional depth: the lines y=±(b/a)x are asymptotes. From the equation, the difference between (b/a)∣x∣ and ∣y∣ is b² divided by their positive sum, which tends to zero as ∣x∣ grows. This justifies approach to those lines; the branches do not terminate where a plotting window stops.