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06_Ex 6.6

Exercise Questions

QuestionsLinks
Q1. Find an equation of the hyperbola with the given…6.6 Q-1
Q2. Find the centre, foci, eccentricity, vertices…6.6 Q-2
Q3. Let 0<a<c,F(c,0)0<\mathbf{a}<\mathbf{c}, \mathbf{F}(-\mathbf{c}, 0) and F(c,0)\mathbf{F}^{\prime}(\mathbf{c}, 0) be two fixed points…6.6 Q-3
Q4. Using Question 3 . Find an equation of hyperbola…6.6 Q-4
Q5. For any point on the hyperbola the difference of…6.6 Q-5
Q6. Two listening posts hear the sound of an enemy gun…6.6 Q-6

Overview

This exercise focuses on understanding and solving problems related to hyperbolas, a fundamental topic in conic sections. Hyperbolas are used in various scientific and real-life applications, such as satellite positioning, astronomy, and acoustics. Through these questions, we explore properties like vertices, foci, directrices, and their equations.


Key Concepts

  1. Definition of Hyperbola: A hyperbola is the locus of points such that the absolute difference of distances from two fixed points (foci) is constant.

  2. Standard Equations:

    • Horizontal Hyperbola:
x2a2y2b2=1 \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
  • Vertical Hyperbola:
y2a2x2b2=1 \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
  1. Components:

    • Vertices: Points where the hyperbola intersects its transverse axis.
    • Foci: Fixed points, each located at a distance cc from the center.
    • Directrices: Lines associated with the eccentricity (ee) of the hyperbola.
  2. Relationships:

    • c2=a2+b2c^2 = a^2 + b^2
    • e=cae = \frac{c}{a}

Important Formulas

  1. Eccentricity:
e=ca,where c=a2+b2. e = \frac{c}{a}, \quad \text{where } c = \sqrt{a^2 + b^2}.
  1. Vertices and Foci:
    • Horizontal Hyperbola:
Vertices: (±a,0),Foci: (±c,0). \text{Vertices: } (\pm a, 0), \quad \text{Foci: } (\pm c, 0).
  • Vertical Hyperbola:
Vertices: (0,±a),Foci: (0,±c). \text{Vertices: } (0, \pm a), \quad \text{Foci: } (0, \pm c).
  1. Directrices:
    • Horizontal Hyperbola:
x=±ae. x = \pm \frac{a}{e}.
  • Vertical Hyperbola:
y=±ae. y = \pm \frac{a}{e}.
  1. General Form: Transformations can adjust the center of the hyperbola from (0,0)(0, 0) to (h,k)(h, k), resulting in:
(xh)2a2(yk)2b2=1. \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.

Tips and Tricks

  1. Identifying the Axis:

    • The term with the positive denominator in the standard equation determines the transverse axis (e.g., x2a2\frac{x^2}{a^2} indicates a horizontal axis).
  2. Symmetry: Hyperbolas are symmetric with respect to both their axes and the center.

  3. Directrix-Eccentricity Relation: Use the eccentricity (ee) to find the directrix, which is perpendicular to the transverse axis.

  4. Graphing: Sketching the hyperbola involves plotting vertices, foci, and asymptotes for guidance.


Summary

  • A hyperbola is defined as the set of points where the difference of distances to two foci is constant.
  • The transverse axis determines the orientation of the hyperbola (horizontal or vertical).
  • The relationship c2=a2+b2c^2 = a^2 + b^2 connects the semi-major axis (aa), semi-minor axis (bb), and the distance to the focus (cc).
  • Key components include vertices, foci, directrices, and eccentricity.
  • Equations can be transformed by changing the center to any arbitrary point (h,k)(h, k).

Reference

By Sir Shahzad Sair: