5.04a - Hyperbola at Origin | High School

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High School Physics
Academic year:

2024
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5.04 The Hyperbola Creating a Hyperbola • Plane intersects both cones • Does not pass through the centre of figure Hyperbola • The locus of a point for which the absolute value of the difference between the distances from two fixed points (foci) is constant. 𝐾 = 𝑃1 𝐹 ′ − 𝑃1 𝐹 = 𝑃2 𝐹 ′ − 𝑃2 𝐹 = ⋯ Vertices Intersection Points with the focal axis CENTRED AT THE ORIGIN Horizontal Hyperbola Rule 2 2 𝑥 𝑦 − = 1 2 2 𝑎 𝑏 Hyperbola with Foci on x-axis (horizontal) http://www.mathwarehouse.com/hyperbola/graph-equation-of-a-hyperbola.php Horizontal Hyperbola Foci The coordinates of the foci are found using 2 2 𝑐 =𝑎 +𝑏 2 For Proof: http://mathworld.wolfram.com/Hyperbola.html Horizontal Hyperbola Summary • Equation of Asymptotes: 𝑏 𝑦=± 𝑥 𝑎 • Coordinates of Foci: 𝑐 2 = 𝑎2 + 𝑏 2 • Rule: – Standard 𝑥2 𝑦2 − 2=1 2 𝑎 𝑏 – General 𝐴𝑥 2 − 𝐵𝑦 2 + 𝐶 = 0 Vertical Hyperbola Rule 2 𝑥 2 𝑦 𝑎 2 𝑏 − 2 = −1 Hyperbola with Foci on y-axis 𝒃 𝐛 − − 𝒙 𝐚𝒂 𝒃 FIX 𝒂 𝒙 (Vertical) Vertical Hyperbola Foci The coordinates of the foci are found from 𝑐 2 2 =𝑎 +𝑏 2 Vertical Hyperbola Summary • Rule of Asymptotes: 𝑏 𝑦= ± 𝑥 𝑎 • Coordinates of Foci 𝑐 2 = 𝑎2 + 𝑏2 • Rule: – Standard 𝑥2 𝑦2 − 2 = −1 2 𝑎 𝑏 – General 𝐴𝑥 2 − 𝐵𝑦 2 + 𝐶 = 0 Finding the Rule 1. Horizontal/Vertical? Choose the rule based on this 2. Find 𝑎 and 𝑏 3. Plug in the numbers Page 355 #1a • HORIZONTAL • =1 • a and b given 𝑎 = 2 and 𝑏 = 5 2 2 𝑥 𝑦 − =1 4 25 Find the Rule #2 • A focal point exists at (0, -5) and a vertex at (0, 4) • VERTICAL • = −1 • b and c given 𝑏 = 4 and 𝑐 = 5 𝑐 2 = 𝑎2 + 𝑏 2 → 𝑎 = 3 2 2 𝑥 𝑦 − = −1 9 16 Find the Rule #3 • A hyperbola has x-intercepts at 𝑥 = ±7 and the focal axis is 2 130 units long • HORIZONTAL • =1 • a and c given 𝑎 = 7 and 𝑐 = 130 𝑐 2 = 𝑎2 + 𝑏 2 → 𝑏 = 9 2 2 𝑥 𝑦 − =1 49 81 Rearranging – Standard to General 𝑥2 𝑦2 − = −1 16 64 Find LCM 𝑥2 𝑦2 64 − = 64 −1 16 64 Apply LCM 4𝑥 2 − 𝑦 2 = −64 Set to Zero 4𝑥 2 − 𝑦 2 + 64 = 0 Try It 2 2 𝑥 𝑦 − =1 100 25 Rearranging – General to Standard 7𝑥 2 − 3𝑦 2 − 84 = 0 Isolate the constant 7𝑥 2 − 3𝑦 2 = 84 Divide by the constant 7𝑥 2 − 3𝑦 2 84 Simplify as much as possible 84 = 84 𝑥2 𝑦2 − =1 12 28 Try It 2 2 3𝑥 − 8𝑦 + 168 = 0 Finding Focal Points • Given Standard Form 𝑥2 𝑦2 − = −1 9 64 • Given General Form 81𝑥 2 − 16𝑦 2 − 1296 = 0 Graphing from a Rule 1. Horizontal/Vertical? 2. Find ‘𝑎’ and ‘𝑏’ 3. Solve for the asymptotes 𝑏 𝑦= ± 𝑥 𝑎 4. Solve for Foci 2 2 2 𝑐 =𝑎 +𝑏 𝑥2 9 • Horizontal • 𝑎 = 3 and 𝑏 = 4 • Asymptotes 4𝑥 𝑦=± 3 • Foci 𝐹 𝑐, 0 𝑎𝑛𝑑 𝐹′ −𝑐, 0 𝑐 2 = 𝑎2 + 𝑏 2 𝑐 2 = 9 + 16 𝑐 2 = 25 𝑐=5 𝐹 5,0 𝑎𝑛𝑑 𝐹′ −5,0 − 𝑦2 16 =1 Inequalities Horizontal ≤ ≥ Vertical 𝑥2 𝑦2 𝑥2 𝑦2 − 2≤1 − 2 ≤ −1 2 2 𝑎 𝑏 𝑎 𝑏 INSIDE OUTSIDE 𝑥2 𝑦2 𝑥2 𝑦2 − 2>1 − 2 > −1 2 2 𝑎 𝑏 𝑎 𝑏 OUTSIDE INSIDE Homework • 5.04a – Hyperbola Centred at the Origin • Visions 2 • P193: 3ace, 4ace, 5ace, 9bd, 11ade