--- title: "Hyperbola: locus and standard equation" subject: "Mathematics I" chapter: 3 key: "semester_1/mathematics_i:hyperbola-standard-equation-locus" source: https://www.bctengineer.com/q/semester_1/mathematics_i/hyperbola-standard-equation-locus site: BCT Engineering --- # Hyperbola: locus and standard equation Mathematics I · Chapter 3: Plane Analytic Geometry - Asked in: 2074 Asoj; 2072 Chaitra; 2071 Chaitra; 2070 Chaitra; 2069 Chaitra; 2067 Asar (asked 6 times) - Video: [HYPERBOLA | Definition, Foci, Vertex, Axes, Latus Rectum of Hyperbola | 2D GEOMETRY - By GP Sir](https://www.youtube.com/watch?v=U6MnDvhQqqw) — Dr.Gajendra Purohit - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/mathematics_i/hyperbola-standard-equation-locus - Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3 ## Question Derive the equation of hyperbola in standard form. (Show that the locus of a point which moves in such a way that the difference of its distances from two fixed points is constant is a hyperbola.) ## Short answer Set foci at (±c,0) and constant difference 2a. From |PF₁-PF₂|=2a, squaring yields x²/a² - y²/b²=1 where b²=c²-a². ## Answer A hyperbola is the locus of a point P that moves so that the difference of its distances from two fixed points (foci) is a constant, 2a, which is less than the distance between the foci. #### Setting up Take the foci F₁(−c, 0) and F₂(c, 0), with the origin at their mid-point and c > a > 0. Let P(x, y) be any point of the locus, so |PF₁ − PF₂| = 2a #### Derivation Take PF₁ − PF₂ = 2a (the other branch gives the same final equation): √((x+c)² + y²) − √((x−c)² + y²) = 2a Move the second root to the right and square: (x+c)² + y² = 4a² + 4a√((x−c)² + y²) + (x−c)² + y² Expanding, the terms x², y², c² cancel and 4cx = 4a² + 4a√((x−c)² + y²), so cx − a² = a√((x−c)² + y²) Square again: c²x² − 2a²cx + a⁴ = a²x² − 2a²cx + a²c² + a²y² (c² − a²)x² − a²y² = a²(c² − a²) Since c > a, put b² = c² − a² (b > 0): b²x² − a²y² = a²b² Divide by a²b²: **x²/a² − y²/b² = 1** This is the standard equation of the hyperbola. #### Properties read from it - Centre (0, 0); vertices (±a, 0); transverse axis 2a, conjugate axis 2b. - Foci (±ae, 0), where b² = a²(e² − 1) and e = c/a > 1. - Directrices x = ±a/e. Asymptotes y = ±(b/a)x. If the foci are on the y-axis, the equation becomes y²/a² − x²/b² = 1. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which defines a hyperbola as a locus? A. Sum of distances from two fixed points is constant B. Difference of distances from two fixed points is a constant 2a, less than the distance between the foci C. Distance from a fixed point equals distance from a fixed line D. Product of distances from two fixed points is constant Answer: B. The hyperbola is the locus with |PF1 − PF2| = 2a < F1F2; the sum version gives an ellipse. (Key point: Definition of the hyperbola) 2. After simplifying, (c^2 − a^2)x^2 − a^2y^2 = a^2(c^2 − a^2). Why can we put b^2 = c^2 − a^2? A. Because c > a, so c^2 − a^2 is positive B. Because c < a, so the quantity is positive C. Because a = b is assumed D. Because c = a for every hyperbola Answer: A. For a hyperbola the foci distance c exceeds a, so c^2 − a^2 > 0 and can be named b^2. (Key point: Why b^2 = c^2 − a^2) 3. How is the root removed in the derivation? A. Differentiate the equation twice B. Substitute x = a cosθ C. Add the two distance equations only once and cancel D. Isolate one root and square, then isolate the remaining root (cx − a^2 = a·root) and square again Answer: D. Two successive squarings eliminate both square roots and give b^2x^2 − a^2y^2 = a^2b^2. (Key point: Derivation step) 4. Which is the standard equation and relation to eccentricity? A. x^2/a^2 + y^2/b^2 = 1 with b^2 = a^2(1 − e^2) B. x^2/a^2 − y^2/b^2 = 1 with b^2 = a^2(1 − e^2) C. x^2/a^2 − y^2/b^2 = 1 with b^2 = a^2(e^2 − 1), e > 1 D. x^2/b^2 − y^2/a^2 = 1 with e < 1 Answer: C. The hyperbola has e = c/a > 1, hence b^2 = c^2 − a^2 = a^2(e^2 − 1). 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