A hyperbola is the set of all points P(x, y) in the plane for which the absolute difference of the distances to two fixed points (the foci) is a positive constant. Place the foci at F₁ = (−c, 0) and F₂ = (c, 0) with c > 0. Let the constant difference be 2a with 0 < a < c. Thus, for any point P(x, y) on the right branch we have:[br]√((x − c)² + y²) − √((x + c)² + y²) = 2a.[br](Using the right branch ensures the left-hand side is nonnegative; the left branch corresponds to the negative of this equation.)[br][br][b]Step 1–Isolate and square once[/b][br]Move one square root to the other side and square both sides: √((x − c)² + y²) = 2a + √((x + c)² + y²)[br]Squaring gives (carefully expanding and simplifying):[br](x − c)² + y² = 4a² + (x + c)² + y² + 4a√((x + c)² + y²).[br]Cancel y² and rearrange: (x − c)² − (x + c)² − 4a² = 4a√((x + c)² + y²).[br]Compute the difference on the left: (x − c)² − (x + c)² = −4cx. Hence −4cx − 4a² = 4a√((x + c)² + y²).[br][br][b]Step 2 – Square a second time[/b][br]Square both sides again and divide by 16 to simplify: (cx + a²)² = a²[(x + c)² + y²].[br]Expand both sides:c²x² + 2a²cx + a⁴ = a²x² + 2a²cx + a²c² + a²y².[br]The mixed terms 2a²cx cancel on both sides, leaving (c² − a²)x² + a⁴ − a²c² = a²y².[br]Factor the constant terms and set b² = c² − a² (> 0). Then a⁴ − a²c² = a²(a² − c²) = −a²b².[br]Thus: b²x² − a²b² = a²y².[br][br][b]Step 3 – Standard form[/b][br]Divide both sides by a²b² (which is positive): (x² / a²) − (y² / b²) = 1, where b² = c² − a² and 0 < a < c. This is the standard (horizontal) hyperbola with transverse axis along the x-axis. Vertices are at (±a, 0) and[br]foci at (±c, 0).[br][br][b]Domain note right/left branches)[/b][br]Because we began with √((x − c)² + y²) − √((x + c)² + y²) = 2a ≥ 0, the algebra corresponds to the right branch, which has x ≥ a. The left branch satisfies the equation with the opposite sign (or, equivalently, the[br]same standard form).[br][br][b]Asymptotes[br][/b]From (x² / a²) − (y² / b²) = 1, rearrange to y² = (b²/a²)(x² − a²). For large |x|, the curve approaches the lines[br]y = ±(b/a) x, which are the asymptotes. (These come directly from setting the right-hand side[br]of the standard form equal to zero.)[br][br][b]Key relations and parameters[/b][br]• Relationship among parameters: c² = a² + b² (so the[br]eccentricity e = c/a > 1).[br]• Distance between vertices: 2a; distance between foci: 2c.[br]• Length of latus rectum: 2b²/a (through each focus).[br]• Vertical/transverse case: swapping x and y yields (y² / a²) − (x² / b²) = 1[br]with foci at (0, ±c).