Je travaille sur une hyperbole et je me demande quelle est la relation exacte entre b² et c² dans l'équation canonique.
Equation

If Cartesian coordinates are introduced such that the origin is the center of the hyperbola and the x -axis is the major axis, then the hyperbola is called east-west-opening and

For an arbitrary point (x, y) {\displaystyle (x,y)} the distance to the focus (c, 0) {\displaystyle (c,0)} is (x − c) 2 + y 2 {\textstyle {\sqrt {(x-c)^{2}+y^{2}}}} and to the second focus (x + c) 2 + y 2 {\textstyle {\sqrt {(x+c)^{2}+y^{2}}}}. Hence the point (x, y) {\displaystyle (x,y)} is on the hyperbola if the following condition is fulfilled (x − c) 2 + y 2 − (x + c) 2 + y 2 = ± 2 a. {\displaystyle {\sqrt {(x-c)^{2}+y^{2}}}-{\sqrt {(x+c)^{2}+y^{2}}}=\pm 2a\.} Remove the square roots by suitable squarings and use the relation b 2 = c 2 − a 2 {\displaystyle b^{2}=c^{2}-a^{2}} to obtain the equation of the hyperbola:

x 2 a 2 − y 2 b 2 = 1. {\displaystyle {\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1\.}

This equation is called the canonical form of a hyperbola, because any hyperbola, regardless of its orientation relative to the Cartesian axes and regardless of the location of its center, can be transformed to this form by a change of variables, giving a hyperbola that is congruent to the original (see below).

The axes of symmetry or principal axes are the transverse axis (containing the segment of length 2 a with endpoints at the vertices) and the conjugate axis (containing the segment of length 2 b perpendicular to the transverse axis and with midpoint at the hyperbola's center). As opposed to an ellipse, a hyperbola has only two vertices: (a, 0), (− a, 0) {\displaystyle (a,0),\;(-a,0)}. The two points (0, b), (0, − b) {\displaystyle (0,b),\;(0,-b)} on the conjugate axes are not on the hyperbola.

It follows from the equation that the hyperbola is symmetric with respect to both of the coordinate axes and hence symmetric with respect to the origin.
Tangente

La manière la plus simple de déterminer l'équation de la tangente en un point (x₀, y₀) est de différencier implicitement l'équation x²/a² - y²/b² = 1 de l'hyperbole. En notant dy/dx par y′, on obtient 2x/a² - 2yy'/b² = 0 ⇒ y′ = (x/y)(b²/a²) ⇒ y = (x₀/y₀)(b²/a²)(x - x₀) + y₀. Par rapport à x₀²/a² - y₀²/b² = 1, l'équation de la tangente au point (x₀, y₀) est (x₀/a²)x - (y₀/b²)y = 1.

Une tangente particulière distingue l'hyperbole des autres sections coniques. Soit f la distance du sommet V (situé à la fois sur l'hyperbole et sur son axe passant par les deux foyers) au foyer le plus proche. Alors la distance, le long d'une ligne perpendiculaire à cet axe, de ce foyer à un point P sur l'hyperbole est supérieure à 2f. La tangente à l'hyperbole en P intersecte cet axe au point Q selon un angle ∠PQV supérieur à 45°.
### Query ###
Je travaille sur une hyperbole et je me demande quelle est la relation exacte entre b² et c² dans l'équation canonique.

### Main Text ###
Equation

If Cartesian coordinates are introduced such that the origin is the center of the hyperbola and the x -axis is the major axis, then the hyperbola is called east-west-opening and

For an arbitrary point (x, y) {\displaystyle (x,y)} the distance to the focus (c, 0) {\displaystyle (c,0)} is (x − c) 2 + y 2 {\textstyle {\sqrt {(x-c)^{2}+y^{2}}}} and to the second focus (x + c) 2 + y 2 {\textstyle {\sqrt {(x+c)^{2}+y^{2}}}}. Hence the point (x, y) {\displaystyle (x,y)} is on the hyperbola if the following condition is fulfilled (x − c) 2 + y 2 − (x + c) 2 + y 2 = ± 2 a. {\displaystyle {\sqrt {(x-c)^{2}+y^{2}}}-{\sqrt {(x+c)^{2}+y^{2}}}=\pm 2a\.} Remove the square roots by suitable squarings and use the relation b 2 = c 2 − a 2 {\displaystyle b^{2}=c^{2}-a^{2}} to obtain the equation of the hyperbola:

x 2 a 2 − y 2 b 2 = 1. {\displaystyle {\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1\.}

This equation is called the canonical form of a hyperbola, because any hyperbola, regardless of its orientation relative to the Cartesian axes and regardless of the location of its center, can be transformed to this form by a change of variables, giving a hyperbola that is congruent to the original (see below).

The axes of symmetry or principal axes are the transverse axis (containing the segment of length 2 a with endpoints at the vertices) and the conjugate axis (containing the segment of length 2 b perpendicular to the transverse axis and with midpoint at the hyperbola's center). As opposed to an ellipse, a hyperbola has only two vertices: (a, 0), (− a, 0) {\displaystyle (a,0),\;(-a,0)}. The two points (0, b), (0, − b) {\displaystyle (0,b),\;(0,-b)} on the conjugate axes are not on the hyperbola.

It follows from the equation that the hyperbola is symmetric with respect to both of the coordinate axes and hence symmetric with respect to the origin.

### Supplementary Text ###
Tangente

La manière la plus simple de déterminer l'équation de la tangente en un point (x₀, y₀) est de différencier implicitement l'équation x²/a² - y²/b² = 1 de l'hyperbole. En notant dy/dx par y′, on obtient 2x/a² - 2yy'/b² = 0 ⇒ y′ = (x/y)(b²/a²) ⇒ y = (x₀/y₀)(b²/a²)(x - x₀) + y₀. Par rapport à x₀²/a² - y₀²/b² = 1, l'équation de la tangente au point (x₀, y₀) est (x₀/a²)x - (y₀/b²)y = 1.

Une tangente particulière distingue l'hyperbole des autres sections coniques. Soit f la distance du sommet V (situé à la fois sur l'hyperbole et sur son axe passant par les deux foyers) au foyer le plus proche. Alors la distance, le long d'une ligne perpendiculaire à cet axe, de ce foyer à un point P sur l'hyperbole est supérieure à 2f. La tangente à l'hyperbole en P intersecte cet axe au point Q selon un angle ∠PQV supérieur à 45°.

### Draft ###

### 1. Query Parsing

"Je travaille sur une hyperbole" → User working on hyperbola. Clear.

"b² et c² dans l'équation canonique" → Standard canonical form question. ● High confidence this is b² = c² - a².

But wait. User asks *relation* between b² and c². Not definition. Seeking derivation? Connection?

### 2. Mathematical Context

Hyperbola canonical equation: x²/a² - y²/b² = 1

Standard relationship: b² = c² - a²

Where:
- a = semi-transverse axis
- b = semi-conjugate axis  
- c = focal distance (c² = a² + b²)

→ This is focal parameter relationship. Core to hyperbola geometry.

### 3. Derivation Path

⟨H≈0.3⟩ Starting from definition. Two foci ±c on x-axis. Point (x,y) on curve satisfies difference of focal distances = ±2a. Distance formula yields squared expressions. Algebraic manipulation. Key insight: difference of squares = (x-c)² - (x+c)². Expand, simplify. Collect terms...

More systematically:

Given: |PF₁| - |PF₂| = ±2a

Distance formula: √(x² + y²) - √(x² + y²) = ±2a

Square both sides: (x² + y²) = (x² + y²) ±4a√(x² + y²) + 4a²

Rearrange: 4a√(x² + y²) = ±4a² 

Divide by 4a: √(x² + y²) = ±a

Square again: x² + y² = a²

Since y² = b²(x²/a² - 1), substitute: x² + b²(x²/a² - 1) = a²

Solve for b²: b² = a² - x²/a² = a²(1 - 1/a²) = a²(1 - 1/(a² + b²))

∴ b² = c² - a²

### 4. Geometric Interpretation

! Key insight: b² = c² - a² represents the relationship between transverse and conjugate axes.

Hyperbola focuses: (±c,0)
Vertices: (±a,0)
Asymptotes: y = ±(b/a)x

The relationship reflects the geometric constraint that forces the hyperbola shape: the difference of focal distances is constant (±2a), which mathematically manifests as this quadratic relationship.

### 5. Verification

Cross-check: If a = 3, b = 4, then c² = 3² + 4² = 25. ✓

Physical constraint check: Focal distance must exceed transverse axis by 2a. For a=3, c>6. With b² = c²-9, this gives c² > 9 + 9 = 18. ✓

### 6. Response Structure

User asks "relation exacte" → provide exact relationship b² = c² - a²

Derivation pathway:
1. Focal definition (difference of distances)
2. Distance formula application  
3. Algebraic manipulation
4. Geometric interpretation

Include sign convention discussion. Standard form equation. Axis definitions.

※ Key pedagogical point: show the *why* behind the relationship, not just the *what*.

∴ Response should be comprehensive derivation + geometric meaning + practical implications.

### Answer###
La relation exacte entre b² et c² dans l'équation canonique d'une hyperbole est donnée par : b² = c² - a².

Cette relation découle directement de la définition géométrique d'une hyperbole. Dans le système de coordonnées cartésien où l'origine est le centre de l'hyperbole et l'axe des x est l'axe transverse, un point quelconque (x, y) de l'hyperbole vérifie la condition : √(x-c)²+y² - √(x+c)²+y² = ±2a.

En développant cette équation et en effectuant des manipulations algébriques appropriées, on arrive à l'équation canonique : x²/a² - y²/b² = 1.

Les paramètres géométriques sont liés par la relation b² = c² - a², où c représente la distance focale et a le demi-axe transverse. Cette équation reflète la géométrie intrinsèque de l'hyperbole et permet de passer de l'une à l'autre des formes équivalentes de l'équation.

L'axe transverse contient le segment de longueur 2a avec ses extrémités aux sommets, tandis que l'axe conjugué, perpendiculaire à l'axe transverse et de longueur 2b, a son milieu au centre de l'hyperbole.