Problem 27 - Precalculus: Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} x^{2}+y^{2}=9 \\ x^{2}-y^{2}=1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{5}, 2), (\sqrt{5}, -2), (-\sqrt{5}, 2), (-\sqrt{5}, -2)\).

Step 1: Analyze the given equations

We have two equations in the system: \( x^2 + y^2 = 9 \) and \( x^2 - y^2 = 1 \). These are equations of a circle and a hyperbola respectively.

Step 2: Add and subtract the equations

Add the two equations to eliminate \( y^2 \): \( x^2 + y^2 + x^2 - y^2 = 9 + 1 \). This simplifies to \( 2x^2 = 10 \), giving \( x^2 = 5 \). Subtract the second from the first to eliminate \( x^2 \): \( x^2 + y^2 - (x^2 - y^2) = 9 - 1 \). This simplifies to \( 2y^2 = 8 \), giving \( y^2 = 4 \).

Step 3: Solve for x and y

Since \( x^2 = 5 \), it follows that \( x = \pm \sqrt{5} \). Similarly, since \( y^2 = 4 \), we have \( y = \pm 2 \).

Step 4: Find all solution pairs

The combinations of \( x = \pm \sqrt{5} \) and \( y = \pm 2 \) give the solution pairs: \((x, y) = (\sqrt{5}, 2), (\sqrt{5}, -2), (-\sqrt{5}, 2), (-\sqrt{5}, -2)\).

Step 5: Verify solutions

Substitute each solution pair back into both equations to verify they satisfy the system. For each pair, check that \( x^2 + y^2 = 9 \) and \( x^2 - y^2 = 1 \). All check out correctly.

Key Concepts

Equation of a Circle

Let's explore what an equation of a circle looks like. The general form of a circle's equation is \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle.
In simpler terms, every point \((x, y)\) that satisfies this equation lies on the circumference of a circle with center at the origin \((0, 0)\) and radius \(r\).
In the equation \(x^2 + y^2 = 9\), you can see that it represents a circle with a radius of 3 units because \(r^2\) is 9, which gives us \(r = 3\).

• Origin Center: The circle is centered at the origin, meaning the center of the circle is at \((0, 0)\).
• Radius: The square root of 9 is 3, so every point on this circle is 3 units away from the origin.

Understanding this foundational structure helps us grasp how circles behave in a coordinate plane.

Equation of a Hyperbola

Hyperbolas might seem more complex due to their unique shape. The simplest form of a hyperbola's equation is \(x^2 - y^2 = c\), where \(c\) is a constant.
This particular equation, \(x^2 - y^2 = 1\), defines a hyperbola centered at the origin. It is oriented along the x and y axes.

• Branches: Unlike an ellipse or a circle, a hyperbola has two separate branches.
• Orientation: This equation has axes as its symmetry, meaning it can open left and right or upward and downward.

In our problem, \(x^2 - y^2 = 1\), signifies it opens along the x-axis providing potential paths where graphs meet.

Solution of Equations

To solve a system of equations means finding all pairs \((x, y)\) that satisfy both equations simultaneously.
The ultimate goal in such systems is to determine the points of intersection on the graph.
By solving these equations, we achieved the pairs that lie on both the circle and the hyperbola simultaneously.

• Focus on Compatibility: Begin by acknowledging the graphical interpretation of each equation like the given circle and hyperbola.
• Intersection Points: We calculated all possible pairs, ensuring that each one fulfills both equations within the system.

The results are the exact pairs where these two distinct shapes cross paths on a cartesian plane.

Elimination Method

Elimination Method comes in handy when dealing with linear equations, by removing one variable to simplify finding solutions.
Interesting fact: It can also effectively address non-linear equations like those we observed in this exercise.
Our approach involved:

• Adding Equations: Directly adding \(x^2 + y^2 = 9\) and \(x^2 - y^2 = 1\) to eliminate overlapping terms and simplify the equation, resulting in \(x^2\) and subsequently finding \(x^2 = 5\).
• Subtracting Equations: Conversely, subtraction helps in eliminating \(x^2\) and simplifies the process for resolving \(y^2 = 4\).

The beauty of the elimination method shines as we seamlessly reduce complexity and find viable solutions to all equations in the system.