Problem 44 - College Algebra with Corequisite Support

Question

For the following exercises, graph the system of inequalities. Label all points of intersection. $$ \begin{array}{l} x^{2}-y^{2}>-4 \\ x^{2}+y^{2}<12 \end{array} $$

Step-by-Step Solution

Answer

Graph the circle and the hyperbola; intersection points are (2, √8), (2, -√8), (-2, √8), (-2, -√8).

Step 1: Understand the Inequalities

We begin by interpreting the inequalities. The first inequality, \(x^2 - y^2 > -4\), is a hyperbola, and the second inequality, \(x^2 + y^2 < 12\), is a circle centered at the origin with radius \(\sqrt{12}\). These inequalities represent regions in the coordinate plane.

Step 2: Rewrite the Inequalities

For clarity, rewrite the system of inequalities as \(x^2 - y^2 = -4\) (hyperbola) and \(x^2 + y^2 = 12\) (circle). These equations are the boundaries of the regions described by our inequalities.

Step 3: Graph the Circle

Plot the circle \(x^2 + y^2 = 12\). This circle has a radius of \(\sqrt{12}\), approximately 3.46. It contains all points within this distance from the origin. The region defined by \(x^2 + y^2 < 12\) is all points inside this circle.

Step 4: Graph the Hyperbola

Plot the hyperbola given by \(x^2 - y^2 = -4\). This curve has two branches, which open vertically. Points above these branches satisfy \(x^2 - y^2 > -4\).

Step 5: Identify Intersection Points

To find intersections, solve the equations \(x^2 - y^2 = -4\) and \(x^2 + y^2 = 12\) simultaneously. Add the equations to eliminate \(y\), giving \(2x^2 = 8\), or \(x^2 = 4\). Hence, \(x = \pm 2\). Substitute \(x = 2\) into one of the equations to find \(y\).

Step 6: Solve for y-values

Substitute \(x = 2\) into \(x^2 + y^2 = 12\): \(4 + y^2 = 12\), so \(y^2 = 8\), thus \(y = \pm \sqrt{8}\). The intersection points are \((2, \sqrt{8})\) and \((2, -\sqrt{8})\).

Step 7: Plot and Label Intersections

On the graph, plot the intersection points: \((2, \sqrt{8})\) and \((2, -\sqrt{8})\), and their labeled equivalents for \(x = -2\). Since both \(x = 2\) and \(x = -2\) satisfy, there are four intersections in total.

Key Concepts

Hyperbola

A hyperbola is a unique type of conic section that appears when a plane cuts through both nappes of a double cone. This results in two disconnected curves, known as the branches of the hyperbola. Each branch of a hyperbola seems symmetrical and opens either horizontally or vertically in typical representations.

The standard form of a hyperbola that is related to the inequality in the exercise is given by the equation \(x^2 - y^2 = C\), where \(C\) is a constant. Here, with \(C = -4\), we're looking at a hyperbola with its center at the origin. The inequality \(x^2 - y^2 > -4\) defines the region exterior to these branches where solutions are valid. These branches stretch out infinitely, so solutions exist all around them but outside the main curves.

When sketching or plotting a hyperbola, start by locating the center and pay attention to its asymptotes, which help guide the exact position of the hyperbola's branches.

Circle

A circle, a fundamental geometric shape, is comprised of all points that are equidistant from a center point. In our exercise, the circle is specified by the equation \(x^2 + y^2 = 12\). This marks a circle centered at the origin, \((0, 0)\), with a radius equal to \(\sqrt{12}\), which is approximately 3.46.

The inequality associated with the circle, \(x^2 + y^2 < 12\), designates the region inside this circle. Thus, any point within a circular area having this radius from the origin will satisfy the inequality.

• Center: Point (0,0).
• Radius: \(\sqrt{12} \approx 3.46\).
• Region defined: Inside the circle.

Understanding these characteristics of a circle is pivotal for successfully graphing and identifying how different figures intersect on a coordinate plane.

Coordinate Plane

The coordinate plane, or Cartesian plane, is a two-dimensional space delineated by the intersection of a horizontal line (x-axis) and a vertical line (y-axis). This grid is utilized for representing points defined by pairs of numbers \((x, y)\), where \(x\) and \(y\) denote the respective horizontal and vertical distances from the origin, which is the point \((0,0)\).

In exercises involving graphing systems of inequalities, like ours with a circle and a hyperbola, the coordinate plane comes into play as the battlefield where we visually plot these geometric figures.

• Axes intersect at the origin \((0,0)\).
• Each point corresponds to \((x, y)\).
• Used for visual representation of equations and solutions.

Proper knowledge of how points and shapes interact in this space is essential for graphing and solving systems of equations or inequalities. The skill to read and interpret plots on a coordinate plane can significantly aid in understanding complex inequalities and intersection points.

Intersection Points

Finding intersection points is a key part of solving systems of equations or inequalities. These are points where the graphs of the equations meet, resulting in common solutions for both equations.

To identify these points, we can solve the equations simultaneously. In our exercise, by handling both the circle's and the hyperbola's equations, we identify points where these figures meet on the coordinate plane. By adding the equations \(x^2 - y^2 = -4\) and \(x^2 + y^2 = 12\), we effectively remove the \(y^2\) terms, simplifying to give \(2x^2 = 8\). This simplifies to \(x^2 = 4\), thus giving real values of \(x\).

Further substituting back, we solve for \(y\) and uncover that the intersection points are two pairs: \((2, \sqrt{8})\) and \((2, -\sqrt{8})\), along with their reflected pairs when \(x = -2\).

• Intersection is where graphs coincide.
• These points satisfy both equations.
• Solutions involve algebraic manipulation and substitution.

This process of solving ensures we grasp where different regions overlap or meet on the coordinate plane, key to understanding how different shapes interact.