Math Problem Statement

Find the equation of the hyperbola with centre at the origin satisfying the following conditions.Find the equation of the hyperbola with centre at the origin satisfying the following conditions.

Solution

To find the equation of a hyperbola with its center at the origin, we generally need certain key pieces of information such as the distance of the foci from the center, the distance of the vertices from the center, or the eccentricity. Since the image you provided didn't specify these details but only mentioned the eccentricity and focus for a circle, I'll outline the general approach depending on what information is typically given:

General Form of a Hyperbola:

A hyperbola centered at the origin can have two standard forms depending on the orientation of the transverse axis:

  1. Horizontal Transverse Axis: x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
  2. Vertical Transverse Axis: y2a2x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

Where:

  • 2a2a is the distance between the vertices,
  • 2c2c is the distance between the foci,
  • b2=c2a2b^2 = c^2 - a^2,
  • e=cae = \frac{c}{a} is the eccentricity.

Steps to Derive the Equation:

  1. Given Foci:

    • If the foci are given at (±c,0)(\pm c, 0) or (0,±c)(0, \pm c), determine whether the hyperbola is horizontal or vertical.
  2. Given Eccentricity (e):

    • If the eccentricity ee and the distance 2a2a (distance between the vertices) are provided, use the relationship: c=aec = ae Then, calculate b2=c2a2b^2 = c^2 - a^2.
  3. Construct the Equation:

    • Substitute a2a^2 and b2b^2 into the appropriate standard form based on the orientation of the hyperbola.

Example:

If we were given that the hyperbola has a center at the origin, vertices at (±5,0)(\pm 5, 0), and eccentricity e=2e = 2, the steps would be:

  1. a=5a = 5 (since the distance from the center to the vertex is 5).
  2. c=ae=5×2=10c = ae = 5 \times 2 = 10.
  3. b2=c2a2=10252=10025=75b^2 = c^2 - a^2 = 10^2 - 5^2 = 100 - 25 = 75.

The equation of the hyperbola is then: x225y275=1\frac{x^2}{25} - \frac{y^2}{75} = 1

If the transverse axis were vertical instead of horizontal, the equation would be: y225x275=1\frac{y^2}{25} - \frac{x^2}{75} = 1


If you have the specific details for the hyperbola you need, please provide them, and I can calculate the exact equation. Otherwise, let me know if you'd like to explore more examples or have any questions.

Related Questions:

  1. How do you determine the orientation of a hyperbola given its equation?
  2. What is the relationship between the semi-major axis, semi-minor axis, and foci of a hyperbola?
  3. How does the eccentricity of a hyperbola affect its shape?
  4. How do you graph a hyperbola given its equation?
  5. What are the asymptotes of a hyperbola, and how are they calculated?

Tip: In hyperbolas, remember that the distance between the foci is always greater than the distance between the vertices. This difference is crucial in identifying and solving hyperbola problems.

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Math Problem Analysis

Mathematical Concepts

Hyperbola
Conic Sections
Eccentricity

Formulas

Standard form of a hyperbola

Theorems

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Suitable Grade Level

High School