Problem 86
Question
Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &\frac{(x-2)^{2}}{36}+\frac{(y+2)^{2}}{25} \leq 1\\\ &\frac{(x+1)^{2}}{9}+\frac{(y-3)^{2}}{25} \leq 1 \end{aligned}$$
Step-by-Step Solution
Verified Answer
Graph both ellipses, then shade the region where their interiors overlap.
1Step 1: Identify the type of conic sections
The given inequalities are in the form of an ellipse, represented by \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) where \((h, k)\) is the center, and \(a\) and \(b\) are the semi-major and semi-minor axes respectively.
2Step 2: Analyze the first ellipse
For the first inequality \( \frac{(x-2)^2}{36} + \frac{(y+2)^2}{25} \leq 1 \), the center is \((2, -2)\). It has a semi-major axis of 6 (because \(\sqrt{36} = 6\), greater than \(\sqrt{25}\)) along the x-axis and a semi-minor axis of 5 along the y-axis.
3Step 3: Analyze the second ellipse
For the second inequality \( \frac{(x+1)^2}{9} + \frac{(y-3)^2}{25} \leq 1 \), the center is \((-1, 3)\). It has a semi-major axis of 5 along the y-axis and a semi-minor axis of 3 along the x-axis.
4Step 4: Graph the first ellipse
On a coordinate plane, plot the center at \((2, -2)\). From this point, draw an ellipse that extends 6 units to the right and left along the x-axis and 5 units up and down along the y-axis. Shade the interior of the ellipse to represent the solution set.
5Step 5: Graph the second ellipse
On the same coordinate plane, plot its center at \((-1, 3)\). From this point, draw an ellipse that extends 3 units to the right and left along the x-axis and 5 units up and down along the y-axis. Shade the interior of this ellipse as well to represent the solution set.
6Step 6: Identify the overlapping region
The overall solution set of the system of inequalities is where the interiors of the two ellipses overlap. This region represents the set of all \((x, y)\) points that satisfy both inequalities.
Key Concepts
EllipsesGraphing SolutionsConic Sections
Ellipses
Ellipses are a fascinating type of conic section. They appear as oval shapes on a plane. They can be cleverly defined through an algebraic equation. In general form, an ellipse is given by the equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \( (h, k) \) represents the center of the ellipse. The values \( a \) and \( b \) denote the lengths of the semi-major and semi-minor axes, respectively.
- The semi-major axis is the longest diameter of the ellipse.
- The semi-minor axis is the shortest diameter.
- If \( a > b \), the ellipse stretches along the x-axis. If \( b > a \), it stretches along the y-axis.
Graphing Solutions
Graphing solutions to systems of inequalities, like the ones involving ellipses, is a hands-on process. It reveals the set of all possible solutions visually on a coordinate plane. The solution is often represented by shading regions. To graph elliptic inequalities, begin by plotting their centers. Use the ellipse equations to decide how to stretch the figure along the axes.
- For example, consider \( \frac{(x-2)^2}{36} + \frac{(y+2)^2}{25} \leq 1 \). The center is \( (2, -2) \), with a semi-major axis of 6 along the x-axis and a semi-minor axis of 5 along the y-axis.
- Next, draw the ellipse and shade the area within to indicate the solution set.
Conic Sections
Conic sections are curves obtained by slicing a cone with a plane. They include circles, ellipses, parabolas, and hyperbolas. Of these, ellipses and circles are closely related as they both signify closed curves.
- Circles are a special type of ellipse where \( a = b \).
- Hyperbolas differ as they split into two distinct open curves.
- Parabolas open in a single direction and appear as U-shaped or inverted-U-shaped curves.
Other exercises in this chapter
Problem 86
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