Problem 37 - Precalculus: Functions and Graphs

Question

Exer \(37-40:\) Find the points of intersection of the graphs of the equations. Sketch both graphs on the same coordinate plane, and show the points of intersection. $$\left\\{\begin{array}{l} x^{2}+4 y^{2}=20 \\ x+2 y=6 \end{array}\right.$$

Step-by-Step Solution

Answer

The points of intersection are (4, 1) and (2, 2).

Step 1: Understand the Equations

The given system consists of two equations: 1) \( x^2 + 4y^2 = 20 \), which represents an ellipse, and 2) \( x + 2y = 6 \), which represents a straight line. Our task is to find the points where these two equations intersect.

Step 2: Express x in Terms of y

Solve the linear equation for \(x\) in terms of \(y\). From \( x + 2y = 6 \), we get \( x = 6 - 2y \). This expression will be substituted into the ellipse equation.

Step 3: Substitute x into the Ellipse Equation

Substitute \( x = 6 - 2y \) into the ellipse equation \( x^2 + 4y^2 = 20 \):\[(6 - 2y)^2 + 4y^2 = 20\] Expand and simplify the expression.

Step 4: Simplify and Solve the Quadratic

Expanding gives \((6 - 2y)^2 = 36 - 24y + 4y^2\). Substitute in the equation:\[36 - 24y + 4y^2 + 4y^2 = 20\] Simplify it to get:\[8y^2 - 24y + 36 = 20\] Further simplify to:\[8y^2 - 24y + 16 = 0\] Divide by 8 to simplify further:\[y^2 - 3y + 2 = 0\] This is the quadratic equation that we solve next.

Step 5: Solve the Quadratic Equation

We solve \( y^2 - 3y + 2 = 0 \) using factorization:\[(y - 1)(y - 2) = 0\] Therefore, \( y = 1 \) and \( y = 2 \).

Step 6: Find Corresponding x Values

For each \( y \) value, substitute back into \( x = 6 - 2y \):- If \( y = 1 \), then \( x = 6 - 2 \times 1 = 4 \).- If \( y = 2 \), then \( x = 6 - 2 \times 2 = 2 \).

Step 7: State the Points of Intersection

The points of intersection are \((4, 1)\) and \((2, 2)\). These are the solutions to the system of equations.

Key Concepts

Ellipse

An ellipse is a smooth, closed curve that resembles a flattened circle and is defined mathematically by the equation \( x^2/a^2 + y^2/b^2 = 1 \). This type of equation describes the set of all points where the sum of the distances from two fixed points, called foci, is constant. The equation within our problem is \( x^2 + 4y^2 = 20 \). You can see this follows the typical form of an ellipse equation, but rearranged slightly.Here, the numerator constants represent the stretches along the x and y axes:

• \( x^2 \) term indicates the horizontal dimension (how wide the ellipse is).
• \( 4y^2 \) term indicates the vertical dimension (how tall the ellipse is), reflecting a greater stretch in the y direction due to the coefficient 4.

By dividing everything by 20, the equation becomes \( (x^2/20) + (y^2/5) = 1 \), showcasing the ellipse's semi-axes. Understanding the structure of this equation is crucial for graphing ellipses and predicting intersections.

Linear Equation

A linear equation is a mathematical statement that creates a straight line when graphed on a coordinate plane. It has a general form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In this problem, the linear equation \( x + 2y = 6 \) represents such a line.To solve for either variable, you perform algebraic operations to rearrange the equation. Here, you can express \( x \) in terms of \( y \):

• Move terms around to isolate \( x \): \( x = 6 - 2y \).

This expression is key for substitution, allowing you to combine the linear equation with the ellipse equation. This combination is fundamental when determining intersection points.

Intersection Points

Intersection points are where two graphs meet on a coordinate plane. For this problem, these points lie on both an ellipse and a line. Finding these points involves combining and solving both the ellipse and the linear equation.The strategy begins with substitution:

• Solve the linear equation for one variable.
• Substitute this expression into the ellipse's equation.

Doing this provides a new equation (often quadratic), which you solve to find specific points shared by both the ellipse and line. In this exercise's solution, the correct intersection points are calculated to be \((4, 1)\) and \((2, 2)\). These points are where the line "pierces" through the ellipse, revealing the precise coordinates where both graphs coincide.

Quadratic Equation

A quadratic equation takes the form \( ax^2 + bx + c = 0 \), involving terms squared to the power of 2. These equations often arise when combining different geometric shapes on a plane, such as an ellipse and a line in our case. Our derived quadratic equation \( y^2 - 3y + 2 = 0 \) originates from substituting the linear equation into the ellipse equation.Solving quadratic equations can be done through various methods:

• Factorization: Express the equation as a product of linear factors. This solution path for \( y^2 - 3y + 2 = 0 \) reveals the roots \( y = 1 \) and \( y = 2 \).
• Quadratic Formula: Use \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for more complex solutions.

In our solution, factorization was appropriate and straightforward, allowing us to pinpoint the intersection points quickly upon substituting back to find corresponding \( x \) values. Quadratic equations are pivotal tools to solve interception problems between curves and straight lines.