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.

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 \( (\sqrt{5}, 0) \) and \( (-\sqrt{5}, 0) \), with a major axis length of \( 2\sqrt{10} \).

For an ellipse, the equation is given by:

\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\]

Where:
- \( a \) is the semi-major axis,
- \( b \) is the semi-minor axis,
- The distance between the center and the foci is \( c \), where \( c^2 = a^2 - b^2 \).

From the given information:
- The distance to the foci \( c = \sqrt{5} \),
- The length of the major axis is \( 2\sqrt{10} \), so the semi-major axis \( a = \sqrt{10} \).

Using the relationship \( c^2 = a^2 - b^2 \):

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

\[
5 = 10 - b^2
\]

\[
b^2 = 5
\]

Thus, the equation of the ellipse is:

\[
\frac{x^2}{10} + \frac{y^2}{5} = 1
\]

### Step 2: Equation of the Hyperbola
The second equation is already given as:

\[
x^2 - y^2 = 1
\]

### Step 3: Solving the System
We now have the system of equations:

1. \( \frac{x^2}{10} + \frac{y^2}{5} = 1 \)
2. \( x^2 - y^2 = 1 \)

We will solve this system by substitution.

First, from the hyperbola equation, solve for \( x^2 \):

\[
x^2 = y^2 + 1
\]

Substitute this expression into the ellipse equation:

\[
\frac{y^2 + 1}{10} + \frac{y^2}{5} = 1
\]

Multiply through by 10 to eliminate the denominators:

\[
(y^2 + 1) + 2y^2 = 10
\]

Simplify:

\[
3y^2 + 1 = 10
\]

\[
3y^2 = 9
\]

\[
y^2 = 3
\]

Thus, \( y = \pm \sqrt{3} \).

### Step 4: Find Corresponding \( x \)-Values
Substitute \( y^2 = 3 \) into the equation \( x^2 = y^2 + 1 \):

\[
x^2 = 3 + 1 = 4
\]

Thus, \( x = \pm 2 \).

### Step 5: Solutions
The solutions to the system are:

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

### Final Answer:
The solution set for the system of equations is:

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

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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.

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 is represented by x^2-y^2=1.

Learn how to solve a system of non-linear equations involving an ellipse and a hyperbola. The ellipse is centered at the origin with foci at (sqrt(5), 0) and (-sqrt(5), 0), and the hyperbola is represented by x^2 - y^2 = 1. Follow a detailed step-by-step approach using algebraic substitution to find the solution. Suitable for Grades 11-12.

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