Math Problem Statement

Solving systems of non-linear equations find the solution of the system of non-linear equation given that the first equation is an ellipse where the center is at the origin and the foci are located at (square root 5, 0) and (negative square root 5, 0) with a major axis of 2 square root 10), and the second equation us represented by x^2-y^2=1.

Solution

To solve the system of nonlinear equations, we first need to deduce the specific form of the equations given the information.

Step 1: Equation of the Ellipse

The problem provides that the ellipse is centered at the origin, and the foci are located at (5,0)(\sqrt{5}, 0) and (5,0)(-\sqrt{5}, 0), with a major axis length of 2102\sqrt{10}.

For an ellipse, the equation is given by:

x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Where:

  • aa is the semi-major axis,
  • bb is the semi-minor axis,
  • The distance between the center and the foci is cc, where c2=a2b2c^2 = a^2 - b^2.

From the given information:

  • The distance to the foci c=5c = \sqrt{5},
  • The length of the major axis is 2102\sqrt{10}, so the semi-major axis a=10a = \sqrt{10}.

Using the relationship c2=a2b2c^2 = a^2 - b^2:

(5)2=(10)2b2(\sqrt{5})^2 = (\sqrt{10})^2 - b^2

5=10b25 = 10 - b^2

b2=5b^2 = 5

Thus, the equation of the ellipse is:

x210+y25=1\frac{x^2}{10} + \frac{y^2}{5} = 1

Step 2: Equation of the Hyperbola

The second equation is already given as:

x2y2=1x^2 - y^2 = 1

Step 3: Solving the System

We now have the system of equations:

  1. x210+y25=1\frac{x^2}{10} + \frac{y^2}{5} = 1
  2. x2y2=1x^2 - y^2 = 1

We will solve this system by substitution.

First, from the hyperbola equation, solve for x2x^2:

x2=y2+1x^2 = y^2 + 1

Substitute this expression into the ellipse equation:

y2+110+y25=1\frac{y^2 + 1}{10} + \frac{y^2}{5} = 1

Multiply through by 10 to eliminate the denominators:

(y2+1)+2y2=10(y^2 + 1) + 2y^2 = 10

Simplify:

3y2+1=103y^2 + 1 = 10

3y2=93y^2 = 9

y2=3y^2 = 3

Thus, y=±3y = \pm \sqrt{3}.

Step 4: Find Corresponding xx-Values

Substitute y2=3y^2 = 3 into the equation x2=y2+1x^2 = y^2 + 1:

x2=3+1=4x^2 = 3 + 1 = 4

Thus, x=±2x = \pm 2.

Step 5: Solutions

The solutions to the system are:

(x,y)=(2,3),(2,3),(2,3),(2,3)(x, y) = (2, \sqrt{3}), (2, -\sqrt{3}), (-2, \sqrt{3}), (-2, -\sqrt{3})

Final Answer:

The solution set for the system of equations is:

(2,3),(2,3),(2,3),(2,3)(2, \sqrt{3}), (2, -\sqrt{3}), (-2, \sqrt{3}), (-2, -\sqrt{3})


Would you like more details on any of the steps?

Related Questions:

  1. How do you derive the standard form of an ellipse from its geometric properties?
  2. What is the significance of the foci in the equation of an ellipse?
  3. How do you solve systems of nonlinear equations graphically?
  4. What are the geometric interpretations of hyperbolas and ellipses?
  5. How do you solve nonlinear systems algebraically by elimination or substitution?

Tip:

When solving systems of nonlinear equations, always check whether substitution or elimination is more efficient, depending on the form of the equations.

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Math Problem Analysis

Mathematical Concepts

Non-linear systems
Ellipse
Hyperbola
Conic sections
Algebraic substitution

Formulas

Equation of an ellipse: x^2/a^2 + y^2/b^2 = 1
Equation of a hyperbola: x^2 - y^2 = 1
Foci formula for ellipse: c^2 = a^2 - b^2

Theorems

Conic sections
Substitution method in solving systems of equations

Suitable Grade Level

Grades 11-12