Problem 32
Question
Exer. 31-32: 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. $$ \left\\{\begin{array}{l} x^{2}+4 y^{2}=36 \\ x^{2}+y^{2}=12 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
Intersection points are (±4, ±2).
1Step 1: Set up the System of Equations
We are given two equations: 1. \( x^{2} + 4y^{2} = 36 \) 2. \( x^{2} + y^{2} = 12 \). The goal is to find the points \((x, y)\) that satisfy both equations simultaneously.
2Step 2: Solve the System Algebraically by Subtraction
Subtract the second equation from the first equation to eliminate \( x^{2} \):\[(x^{2} + 4y^{2}) - (x^{2} + y^{2}) = 36 - 12 \4y^{2} - y^{2} = 24 \3y^{2} = 24\]Solve for \( y^{2} \) by dividing both sides by 3:\[ y^{2} = 8 \]Take the square root of both sides:\[ y = ext{either } oxed{ ext{positive or negative }rac{2 ext{ } imes ext{ } ext{sqrt(2)}}{2}} \].
Key Concepts
Intersection PointsCoordinate GeometryAlgebraic SolutionsEllipses
Intersection Points
When working with a set of equations, one of the main goals is often to find the points where the graphs of the equations meet, known as intersection points. These points are significant because they satisfy both equations simultaneously.
To find intersection points, start by solving the system of equations simultaneously. In our case, the equations are:
Understanding the concept of intersection points not only helps us in mathematics, but it's also a key concept in fields like physics and engineering, providing solutions to real-world intersection problems.
To find intersection points, start by solving the system of equations simultaneously. In our case, the equations are:
- \( x^2 + 4y^2 = 36 \)
- \( x^2 + y^2 = 12 \)
Understanding the concept of intersection points not only helps us in mathematics, but it's also a key concept in fields like physics and engineering, providing solutions to real-world intersection problems.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, allows us to study geometry using a coordinate plane. This powerful tool helps us visualize abstract algebraic equations and their relationships. In this exercise, the coordinate plane becomes a canvas where shapes take form, and their intersections are painted as points of interest.
Both equations provided,
Skills in coordinate geometry are essential, not only for solving textbook problems but also for creating designs in computer graphics and in fields such as architecture and engineering. In various applications, it allows precise modeling of locations, distances, and areas.
Both equations provided,
- \( x^2 + 4y^2 = 36 \)
- \( x^2 + y^2 = 12 \)
Skills in coordinate geometry are essential, not only for solving textbook problems but also for creating designs in computer graphics and in fields such as architecture and engineering. In various applications, it allows precise modeling of locations, distances, and areas.
Algebraic Solutions
Algebraic solutions involve manipulating equations to find the desired values. This process includes operations such as substitution, elimination, and factoring. Solving systems of equations algebraically is a critical skill in mathematics as it bypasses the need for graphical methods, instead relying on pure computation.
In our system of equations, subtraction is used to eliminate \( x^2 \), leaving us with:
Mastering algebraic techniques is fundamental for students as they progress to higher levels of mathematics, where these methods become indispensable tools for tackling more intricate equations, such as those involving multiple variables or higher-degree polynomials.
In our system of equations, subtraction is used to eliminate \( x^2 \), leaving us with:
- \( 4y^2 - y^2 = 24 \)
- \( 3y^2 = 24 \)
Mastering algebraic techniques is fundamental for students as they progress to higher levels of mathematics, where these methods become indispensable tools for tackling more intricate equations, such as those involving multiple variables or higher-degree polynomials.
Ellipses
Ellipses are fascinating geometric figures that look like elongated circles. The equation \( x^2 + 4y^2 = 36 \) describes an ellipse when visualized on a coordinate plane. Unlike circles that have a single radius, ellipses have two radii: the semi-major and semi-minor axes.
For the given equation, the ellipse is centered at the origin \((0,0)\). This is typically the case when the equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, the denominator of \( x^2 \) and \( y^2 \) reflects the relative lengths of each axis. The presence of 4 in \( 4y^2 \) implies that the ellipse is stretched along the y-axis.
Ellipses are not just mathematical curiosities; they have applications in physics (such as planetary orbits) and engineering (like gear design). Knowing how to identify and work with ellipses is crucial for students in understanding more complex, real-world dynamic systems.
For the given equation, the ellipse is centered at the origin \((0,0)\). This is typically the case when the equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, the denominator of \( x^2 \) and \( y^2 \) reflects the relative lengths of each axis. The presence of 4 in \( 4y^2 \) implies that the ellipse is stretched along the y-axis.
Ellipses are not just mathematical curiosities; they have applications in physics (such as planetary orbits) and engineering (like gear design). Knowing how to identify and work with ellipses is crucial for students in understanding more complex, real-world dynamic systems.
Other exercises in this chapter
Problem 32
Exer. 27-44: Find an equation in \(x\) and \(y\) that has the same graph as the polar equation. Use it to help sketch the graph in an \(r \theta\)-plane. $$ r=4
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Exer. 33-34: Find a polar equation of the parabola with focus at the pole and the given vertex. $$ V\left(4, \frac{\pi}{2}\right) $$
View solution Problem 33
Identify the graph of the equation as a parabola (with vertical or horizontal axis), circle, ellipse, or hyperbola. $$\frac{1}{3}(x+2)=y^{2}$$
View solution