Problem 64 - Algebra and Trigonometry

Question

In Exercises \(61-66,\) find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. $$ \left\\{\begin{aligned} \frac{x^{2}}{4}+\frac{y^{2}}{36} &=1 \\ x &=-2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution set to the system of equations is \{(-2, 6), (-2, -6)\}

Step 1: Plot the Linear Equation

Begin by graphing the easy linear equation \(x = -2\). This is a vertical line passing through the point (-2, 0) on the x-axis.

Step 2: Plot the Ellipse

Next, graph the equation of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{36} = 1\). It's centered at the origin (0, 0), with a length of the major axis as 12 (2*6), and of the minor axis as 4 (2*2). It extends along the x-axis by 2 units and along the y-axis by 6 units on both sides of the origin.

Step 3: Identify the Points of Intersection

Observe where the line \(x = -2\) intersects the ellipse. These points are the solutions of the system. From the graph, it's apparent that there are two points of intersection because the vertical line crosses the ellipse at two points.

Step 4: Calculate the Points of Intersection

Substitute \(x = -2\) in the equation of the ellipse to find the y-coordinates. This gives \(\frac{(-2)^{2}}{4}+\frac{y^{2}}{36} = 1\). Solving for \(y\), we find the solutions \(y = 6\) and \(y = -6\). Thus, the points of intersection are (-2, 6) and (-2, -6).

Step 5: Check the solutions

Finally, substitute these solution points into both original equations to verify their validity. If each pair satisfies both equations, then they are indeed the solutions of this system of equations.

Key Concepts

Graphing

Graphing a system of equations involves plotting each equation on the same coordinate system to visually find solutions where they intersect. When two graphs meet at a point, that point is a solution to the system. This method helps visually understand the relationship between equations and the possible solutions they share.
For this exercise, the two equations are a linear equation and an ellipse equation. By graphing both on the same axes, we can easily locate where they intersect.

Ellipse

An ellipse is a set of all points such that the sum of the distances from two fixed points (foci) is constant. In the general equation of an ellipse, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes.
In this problem, we have \(\frac{x^2}{4} + \frac{y^2}{36} = 1\), indicating:

• The center of the ellipse is at the origin \((0,0)\).
• The semi-major axis along the y-axis is 6 (since 36 is larger than 4).
• The semi-minor axis along the x-axis is 2.

This ellipse stretches vertically more than horizontally, forming an oval shape around its center.

Linear Equation

A linear equation represents a straight line on the graph. It has the general form: \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. In this case, the linear equation is\(x = -2\), a vertical line through \((-2, 0)\).
Unlike horizontal or sloped lines, vertical lines have an undefined slope and run parallel to the y-axis. Plotting this is simple: just draw a straight line through \(-2\) on the x-axis, extending both upwards and downwards.

Intersection Points

Intersection points occur where two graphs meet. These are the solution points for the system of equations. For a vertical line intersecting an ellipse, we substitute the line's equation into the ellipse's equation to find these points:
Given \(x = -2\) and\(\frac{x^2}{4} + \frac{y^2}{36} = 1\):

• Substitute \(x = -2\) into the ellipse equation: \(\frac{(-2)^2}{4} + \frac{y^2}{36} = 1\)
• This simplifies to \(\frac{4}{4} + \frac{y^2}{36} = 1\), or \(\frac{y^2}{36} = 0\)
• Solve for \(y\), giving two solutions: \(y = 6\) and \(y = -6\)

Hence, the intersection points are \((-2, 6)\) and \((-2, -6)\). Verifying these by substituting back into the original equations confirms they satisfy both.