Tangents, Normals and Confocal Conics
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Direct answer
Two conics that share foci cannot cross at an arbitrary angle: an ellipse and a hyperbola of the same confocal family always meet orthogonally — the tangent of one is the normal of the other. The family is x²/(a² + λ) + y²/(b² + λ) = 1 with a > b fixed and λ varying; the foci (±√(a² − b²), 0) never move because a² + λ − (b² + λ) = a² − b² is constant. Members with λ > −b² are ellipses, those with −a² < λ < −b² are hyperbolas, and through any point off the axes passes exactly one of each, cutting at right angles.
What you must remember
- Confocal family: x²/(a²+λ) + y²/(b²+λ) = 1 with foci (±c, 0), c² = a² − b² fixed for every λ.
- Type by λ: λ > −b² gives an ellipse; −a² < λ < −b² gives a hyperbola; λ = −b² degenerates.
- Two members per point: through (x₁, y₁) with y₁ ≠ 0, λ satisfies a quadratic with one root above −b² and one between −a² and −b² — one ellipse, one hyperbola.
- Orthogonal intersection: at a common point the product of the two tangent slopes is −1; each curve's tangent is the other's normal.
- Ellipse tangent-normal: tangent xx₁/a² + yy₁/b² = 1; normal a²x/x₁ − b²y/y₁ = a² − b².
- Hyperbola tangent-normal: tangent xx₁/a² − yy₁/b² = 1; normal a²x/x₁ + b²y/y₁ = a² + b².
- Do not merge theorems: confocal orthogonality concerns two different curves crossing; the director circle x² + y² = a² + b² concerns two tangents drawn from one point — different statements with different proofs.
Why confocal conics cross at right angles
Let P(x₁, y₁) lie on both x²/(a²+λ₁) + y²/(b²+λ₁) = 1 and x²/(a²+λ₂) + y²/(b²+λ₂) = 1, with λ₁ > λ₂. Implicit differentiation on the member with parameter λ gives the tangent slope m = −(b²+λ)x₁/[(a²+λ)y₁], so the product of the two slopes is m₁m₂ = x₁²(b²+λ₁)(b²+λ₂)/[y₁²(a²+λ₁)(a²+λ₂)]. Orthogonality demands m₁m₂ = −1, i.e. x₁²(b²+λ₁)(b²+λ₂) + y₁²(a²+λ₁)(a²+λ₂) = 0.
That identity is exactly what subtracting the two conic equations produces: the difference reads x₁²[1/(a²+λ₁) − 1/(a²+λ₂)] + y₁²[1/(b²+λ₁) − 1/(b²+λ₂)] = 0; multiplying through by (a²+λ₁)(a²+λ₂)(b²+λ₁)(b²+λ₂)/(λ₂ − λ₁) delivers the orthogonality condition verbatim. The argument never used which member is the ellipse, so the result is symmetric — wherever a shared-focus ellipse and hyperbola cross, the crossing is a right angle, with the hyperbola's tangent running along the ellipse's normal.
How Advanced frames it
Advanced phrases the theorem as a proof item ("show that the ellipse and hyperbola through a given point with the same foci intersect orthogonally") or feeds it into locus problems; Main stays with tangent-normal equations and with identifying a conic from its λ. The traps: asserting λ > −b² always, ignoring the hyperbola branch below −b²; reading the family as having fixed axes — only the foci are fixed, since a² + λ and b² + λ both drift with λ; and confusing the two perpendicularity results, confocal orthogonality versus the director circle, which describe different configurations entirely. When a proof is demanded, the subtraction step above is the expected line: reproduce it as an identity between slopes rather than asserting the conclusion.
Frequently asked questions
What stays fixed in the confocal family x²/(a²+λ) + y²/(b²+λ) = 1?
Only the foci (±√(a² − b²), 0), since a² + λ − (b² + λ) = a² − b² is independent of λ.
When is a member of the family an ellipse or a hyperbola?
λ > −b² gives an ellipse; −a² < λ < −b² gives a hyperbola.
What does "confocal conics intersect orthogonally" mean?
At their common point the product of tangent slopes is −1, so each curve's tangent is the other's normal.
How many conics of a confocal family pass through a given point?
Two — one ellipse and one hyperbola — for any point not on the coordinate axes.
What is the normal to the ellipse x²/a² + y²/b² = 1 at (x₁, y₁)?
a²x/x₁ − b²y/y₁ = a² − b², the line through (x₁, y₁) with direction (x₁/a², y₁/b²).