Problem 44 - College Algebra

Question

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

The intersection points are \((2, 2\sqrt{2})\), \((2, -2\sqrt{2})\), \((-2, 2\sqrt{2})\), and \((-2, -2\sqrt{2})\).

Step 1: Rewrite and Analyze the Inequalities

Start by analyzing the given inequalities. The inequalities are in the form of conics. The first inequality, \(x^2 - y^2 > -4\), is a hyperbola and can be rewritten as \(x^2 - y^2 + 4 > 0\). The second inequality, \(x^2 + y^2 < 12\), represents a circle centered at the origin with a radius of \(\sqrt{12}\,\) which simplifies to \(2\sqrt{3}\).

Step 2: Determine Intersections of the Equations

To find the intersections of the boundaries of regions defined by these inequalities, solve the equations where the inequalities are equalities: \(x^2 - y^2 = -4\) and \(x^2 + y^2 = 12\). Equate both expressions to solve for \(x\) and \(y\). Set \(x^2 - y^2 = -4\) equal to \(x^2 + y^2 = 12\). Solve the system:\[x^2 - y^2 = -4 \x^2 + y^2 = 12 \]

Step 3: Solve the System of Equations

By adding the two equations:\[x^2 - y^2 + x^2 + y^2 = -4 + 12 \]This simplifies to:\[2x^2 = 8 \x^2 = 4 \x = \pm 2\]Substitute back \(x = 2\) into the equation \(x^2 + y^2 = 12\):\[2^2 + y^2 = 12 \4 + y^2 = 12 \y^2 = 8 \y = \pm 2\sqrt{2}\]The solutions are the points \((2, 2\sqrt{2})\), \((2, -2\sqrt{2})\), \((-2, 2\sqrt{2})\), and \((-2, -2\sqrt{2})\).

Step 4: Graph the System of Inequalities

Graph the two boundaries: the hyperbola \(x^2 - y^2 = -4\) and the circle \(x^2 + y^2 = 12\). Shade the region between the hyperbola and inside the circle where both original inequalities hold. Check points within this region to ensure they satisfy both inequalities, confirming correct shading.

Step 5: Verify Intersection Points

Check the intersection points \((2, 2\sqrt{2})\), \((2, -2\sqrt{2})\), \((-2, 2\sqrt{2})\), and \((-2, -2\sqrt{2})\) to verify these satisfy both boundary equations, ensuring correct labeling on the graph.

Key Concepts

Conic Sections

A conic section is a curve obtained by slicing a double cone with a plane. Conic sections include circles, ellipses, parabolas, and hyperbolas. The different shapes result from the angle at which the plane slices the cone.

In our exercise, we have two conic sections: a hyperbola from the inequality \(x^2 - y^2 > -4\) and a circle described by \(x^2 + y^2 < 12\). These inequalities mean we are looking at portions of these conics rather than their entire shapes originally expressed in equations.

• The circle is centered at the origin \((0,0)\) and has a radius of \(2\sqrt{3}\). It represents all points \((x, y)\) within this circular boundary.
• The hyperbola, in its standard form, displays two diverging "arms" of the graph, which are represented as the regions where \(x^2 - y^2 > -4\).

These types of curves are foundational in understanding geometry and can describe physical phenomena like orbits and optics.

Intersection Points

Intersection points are crucial when dealing with systems of equations or inequalities, as they show where those systems meet.

In our specific problem, we need to find where the hyperbola \(x^2 - y^2 = -4\) intersects the circle \(x^2 + y^2 = 12\). This involves solving a system of equations to find common solutions for \(x\) and \(y\).

Here's how it was done:

• By adding and equating terms from both conical equations, we deduced that \(2x^2 = 8\), which gives \(x = \pm 2\).
• Substituting back for \(x\) allowed us to solve for \(y\), resulting in \(y = \pm 2\sqrt{2}\).

The intersection points are \((2, 2\sqrt{2})\), \((2, -2\sqrt{2})\), \((-2, 2\sqrt{2})\), and \((-2, -2\sqrt{2})\). These points are where both conic sections share a common solution, a critical aspect in graphing systems of inequalities.

Graphing Inequalities

Graphing inequalities involves intersecting regions of solutions for complex, multi-equation systems. It visualizes solutions where multiple conditions are satisfied simultaneously.

There are steps in graphing a system of inequalities like the one we have, which include:

• First sketching the boundary curves: Here, the hyperbola \(x^2 - y^2 = -4\) and the circle \(x^2 + y^2 = 12\). These are the "lines" you'd start with.
• Once graphed, identify where to shade, which represents the solution region. For the initial problem, we shade where both conditions \(x^2 - y^2 > -4\) and \(x^2 + y^2 < 12\) overlap.
• Label all key points, particularly intersection points, as they indicate precise solutions where these inequalities overlap.

Always test a point within the shaded region to ensure it satisfies all inequalities. This step is vital to verifying if you've graphed the inequalities properly. Graphing in this manner helps clarify and solve multi-variable conditions.