Problem 67 - College Algebra with Modeling and Visualization

Question

Solve the system of equations. Give graphical support by making a sketch. $$\begin{aligned} &4 x^{2}+16 y^{2}=64\\\ &x^{2}+y^{2}=9 \end{aligned} $$

Step-by-Step Solution

Answer

The system has four intersection points: \(\left(\pm\sqrt{\dfrac{20}{3}}, \pm\sqrt{\dfrac{7}{3}}\right)\).

Step 1: Analyze the First Equation

The first equation is \(4x^2 + 16y^2 = 64\). Divide every term by 64 to rewrite it in the form \(\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1\). This represents an ellipse centered at the origin with semi-major axis 4 along the x-axis and semi-minor axis 2 along the y-axis.

Step 2: Analyze the Second Equation

The second equation is \(x^2 + y^2 = 9\), which is a circle centered at the origin with a radius of 3.

Step 3: Sketch the Graphs

Draw the two conic sections on the same coordinate plane. The ellipse is wider horizontally, stretching from -4 to 4 on the x-axis and from -2 to 2 on the y-axis. The circle goes out to 3 units in all directions from the origin. The points where these two graphs intersect are the solutions to the system of equations.

Step 4: Find the Intersection Points

Substitute \(x^2 = 9 - y^2\) from the circle's equation into the ellipse's equation, yielding \(4(9-y^2) + 16y^2 = 64\). Simplifying, you get \(36 + 12y^2 = 64\), leading to \(12y^2 = 28\) or \(y^2 = \dfrac{7}{3}\). Solving for \(x^2\), get \(x^2 = 9 - \dfrac{7}{3} = \dfrac{20}{3}\).

Step 5: Calculate Valid Points

The possible combinations of solutions for \(x\) and \(y\) can be found by solving separately for each combination of \(\pm\sqrt{\dfrac{20}{3}}\) and \(\pm\sqrt{\dfrac{7}{3}}\). Hence the four points of intersection on the graph are \(\left( \sqrt{\dfrac{20}{3}}, \sqrt{\dfrac{7}{3}} \right)\), \(\left( -\sqrt{\dfrac{20}{3}}, \sqrt{\dfrac{7}{3}} \right)\), \(\left( \sqrt{\dfrac{20}{3}}, -\sqrt{\dfrac{7}{3}} \right)\), and \(\left( -\sqrt{\dfrac{20}{3}}, -\sqrt{\dfrac{7}{3}} \right)\).

Key Concepts

Ellipse

An ellipse is a shape that looks like a flattened circle. In mathematics, an ellipse can be defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This equation represents an ellipse centered at the origin

• \(a\) is the length of the semi-major axis, which is the longest diameter of the ellipse.
• \(b\) is the length of the semi-minor axis, the shortest diameter of the ellipse.

In the given problem, the equation \(4x^2 + 16y^2 = 64\) simplifies to \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), identifying an ellipse with a semi-major axis of 4 units along the x-axis, and a semi-minor axis of 2 units along the y-axis. Visualizing and understanding the shape and size of this ellipse are crucial in solving systems that involve ellipses.

Circle

A circle is a simple shape and is very symmetrical. It is defined in a plane by the equation \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle. This shape retains consistent distance \(r\) from a central point at all angles.

• The radius \(r\) is the distance from the center of the circle to any point on the circumference.

In our system of equations, the equation \(x^2 + y^2 = 9\) is a circle centered at the origin with a radius of 3 units. Drawing this circle alongside other shapes like ellipses helps in visualizing how these systems behave and intersect.

Intersection Points

Finding the intersection points of a system of equations involves locating the coordinates where two or more graphs cross. These intersection points are the solutions to the system. When dealing with equations like an ellipse and a circle, we substitute and solve to find where they meet

• Solving the system starts by expressing one variable in terms of the other using one of the equations.
• Substituting this expression into the second equation allows solving a single equation with one variable.

For the ellipse and circle \[4x^2 + 16y^2 = 64\]\[x^2 + y^2 = 9\]Substitute \(x^2 = 9 - y^2\) into the ellipse, and solve to get \(y^2 = \frac{7}{3}\) and \(x^2 = \frac{20}{3}\). Thus, you find the intersection points by considering both positive and negative square roots of these results.

Graphical Solution

A graphical solution involves solving equations by drawing them on a coordinate plane and identifying where they intersect. For systems of equations, this method is particularly useful as it offers a visual representation of how the equations relate to each other.

• Each equation is represented as a curve or a line on the graph.
• By plotting these equations, we can see where their paths cross.

In the context of the given system of equations - one being an ellipse and the other a circle - sketching both on the same graph lets you locate their intersection points visually. This step enriches the understanding of how these mathematical curves interact, and verifies the intersection points (solutions) obtained algebraically by looking at the graph. This approach not only aids in checking your algebraic solution but also develops a deeper intuitive understanding of the equations' behaviors.