Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the points of intersection of \(x^{2}+4 y^{2}=20\) and \(x+2 y=6\)

Step-by-Step Solution

Answer

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

Step 1: Identify Equations

The problem provides two equations. The first equation is a conic section: \(x^2 + 4y^2 = 20\), which is an ellipse. The second equation is linear: \(x + 2y = 6\). Our task is to find the points where these two curves intersect.

Step 2: Solve the Linear Equation for x

Start by isolating \(x\) in the second equation \(x + 2y = 6\). Solve for \(x\) to express it in terms of \(y\):\[x = 6 - 2y\].

Step 3: Substitute into the Quadratic Equation

Replace \(x\) in the first equation \(x^2 + 4y^2 = 20\) with the expression found in Step 2, \(6 - 2y\). This gives us:\[(6 - 2y)^2 + 4y^2 = 20\].

Step 4: Expand and Simplify

Expand \((6 - 2y)^2\) to get \(36 - 24y + 4y^2\). Substitute this back into the equation to get:\[36 - 24y + 4y^2 + 4y^2 = 20\]. Combine like terms to obtain:\[8y^2 - 24y + 36 = 20\].

Step 5: Rearrange into a Standard Quadratic Equation

Subtract 20 from both sides of the equation to get:\[8y^2 - 24y + 16 = 0\]. Divide the entire equation by 8 to simplify:\[y^2 - 3y + 2 = 0\].

Step 6: Factor the Quadratic

The equation \(y^2 - 3y + 2 = 0\) can be factored into:\[(y - 1)(y - 2) = 0\]. This gives two potential solutions for \(y\), which are \(y = 1\) and \(y = 2\).

Step 7: Find Corresponding x-values

Use the expression \(x = 6 - 2y\) to find the corresponding \(x\) values for each \(y\).For \(y = 1\), \(x = 6 - 2(1) = 4\).For \(y = 2\), \(x = 6 - 2(2) = 2\).

Step 8: Intersection Points

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

Key Concepts

Ellipse Equations

An ellipse is a type of conic section, which is a curve obtained by intersecting a cone with a plane. The standard form of an ellipse equation is \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\], where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. When these axes are equally distributed around the origin, they define the size and shape of the ellipse.

• The equation given in the exercise is \(x^2 + 4y^2 = 20\). By dividing everything by 20, it can be rewritten as \(\frac{x^2}{20} + \frac{y^2}{5} = 1\).
• This indicates an ellipse with a semi-major axis of \(\sqrt{20}\) and a semi-minor axis of \(\sqrt{5}\). The longer axis is along the x-direction, indicating the ellipse is wider than it is tall.

Ellipses often appear in physics and astronomy, for example, describing orbits of planets or satellites.

Linear Equations

Linear equations represent straight lines. They can be expressed in many forms, with the most common being the slope-intercept form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.In this exercise, the linear equation \(x + 2y = 6\) is given. It provides a direct, simple relationship between \(x\) and \(y\). This can also be rearranged to the form \(y = -\frac{1}{2}x + 3\), showing:

• A slope of -\(\frac{1}{2}\), indicating that for every 1 unit increase in \(x\), \(y\) decreases by \(0.5\).
• The y-intercept is 3, meaning the line crosses the y-axis at (0,3).

This straight line intersects the ellipse at two points, requiring the manipulation and substitution as outlined in the solution to solve for the intersection points.

Quadratic Factoring

Quadratic factoring involves expressing a quadratic equation as a product of its factors. For a quadratic in the form \(ax^2 + bx + c = 0\), factoring can simplify solving. In this context, the equation \(y^2 - 3y + 2 = 0\) was obtained after rearranging and simplifying from the intersecting ellipse and line equations. The factors are:

• \((y - 1)(y - 2) = 0\), indicating solutions \(y = 1\) and \(y = 2\).
• These factors provide potential values for \(y\) that satisfy both initial equations simultaneously.

Factoring simplifies the process of finding solutions by breaking down the equation into more workable parts, making it easier to solve by inspection or using the zero product property.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. They include shapes such as circles, ellipses, parabolas, and hyperbolas. Each conic section has a distinct set of properties and equations that describe their curves.

• Circles are a special type of ellipse with equal semi-major and semi-minor axes.
• Ellipses result when the intersection of the plane and the cone is such that it completely passes through one nap but not the other.
• Parabolas are formed when the plane is parallel to one of the generating lines of the cone.
• Hyperbolas result when the plane intersects both naps.

Understanding conic sections is crucial in a wide range of applications, from engineering to physics, where they help describe system behaviors and optimize designs.