Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. \(25 x^{2}+9 y^{2}+150 x-18 y+9=0\)

Step-by-Step Solution

Answer

The graph is an ellipse centered at (-3, 1) with axes lengths 5 (vertical) and 3 (horizontal).

Step 1: Rearrange the equation

Start with the given equation: \(25 x^{2}+9 y^{2}+150 x-18 y+9=0\). Rearrange the equation by organizing the terms related to \(x\) and \(y\) together: \(25x^{2} + 150x + 9y^{2} - 18y = -9\).

Step 2: Complete the square for x-terms

Focus on the \(x\)-terms: \(25x^{2} + 150x\). Factor out \(25\) to make it easier to complete the square: \(25(x^{2} + 6x)\). Then, complete the square by adding and subtracting \(9\) inside the parentheses: \(25(x^{2} + 6x + 9 - 9) = 25((x+3)^{2} - 9)\).

Step 3: Complete the square for y-terms

For the \(y\)-terms: \(9y^{2} - 18y\), factor \(9\) out: \(9(y^{2} - 2y)\). Complete the square by adding and subtracting \(1\): \(9(y^{2} - 2y + 1 - 1) = 9((y-1)^{2} - 1)\).

Step 4: Simplify and finalize the equation

Substitute the completed squares back into the equation: \(25((x+3)^{2} - 9) + 9((y-1)^{2} - 1) = -9\). Simplify the equation: \(25(x+3)^{2} - 225 + 9(y-1)^{2} - 9 = -9\). Combine similar terms to get \(25(x+3)^{2} + 9(y-1)^{2} = 225\).

Step 5: Identify the conic section

The equation \(\frac{(x+3)^{2}}{9} + \frac{(y-1)^{2}}{25} = 1\) represents an ellipse, where \(a^{2} = 25\) and \(b^{2} = 9\). The ellipse is centered at \((-3, 1)\).

Step 6: Sketch the graph

Using the center \((-3, 1)\) and knowing that the semi-major axis is vertical with length \(5\) and the semi-minor axis is horizontal with length \(3\), sketch the ellipse on the coordinate system.

Key Concepts

Conic Sections

Conic sections are the curves obtained by slicing a cone with a plane. They include circles, ellipses, parabolas, and hyperbolas. Each type has a distinct equation that characterizes its shape and properties.

• **Circle**: All points are equidistant from a center point.
• **Ellipse**: An elongated circle, with two focal points.
• **Parabola**: A curve that is symmetric around a line, with a single focus and directrix.
• **Hyperbola**: Consists of two separate curves, with two focal points.

In the given equation, once rearranged and simplified, it represents an **ellipse**, identified by its standard form. An equation of the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) describes an ellipse, where \((h, k)\) is the center. By completing the solution, it becomes clear this equation matches the form of an ellipse.

Completing the Square

Completing the square is a method used to convert a quadratic equation into a more manageable form, often a perfect square trinomial. This technique is particularly useful in deriving the equations of conic sections. Here's a brief guide on completing the square:

• Factor out the coefficient of the squared term, if necessary.
• Add and subtract the square of half the linear term's coefficient (ensuring balance).
• Rewrite the expression as a binomial square.

For example, in the given equation, separate steps are taken:1. For the \(x\) terms, it becomes \(25((x+3)^2 - 9)\).2. For the \(y\) terms, it results in \(9((y-1)^2 - 1)\).
This method helps express quadratic terms in a square form, which simplifies identifying and graphing conic sections.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to understand shapes, sizes, and relative positions in space using algebraic equations. This field combines algebra with classical geometry.Key aspects include:

• Using coordinate points, plane systems (like Cartesian coordinates).
• Understanding geometric shapes via algebraic equations.
• Solving spatial problems using formulas and properties of figures.

In the context of the exercise, coordinate geometry is crucial:- By converting the given equation into a recognizable conic section (ellipse), the geometric properties such as center (\((-3, 1)\)), axes lengths, and orientation are determined.- It enables sketching the graph accurately by translating algebraic solutions into spatial visualization.

Graph Sketching

Graph sketching involves plotting the visual representation of equations on a coordinate plane, which helps to concretize abstract algebraic concepts. Steps for effective graph sketching:

• Identify the type and key features of the conic section from the equation.
• Determine important points, such as centers, vertices, and axes.
• Use symmetry and scaling to accurately draw the figure.

For the specific case of the ellipse \(\frac{(x+3)^2}{9} + \frac{(y-1)^2}{25} = 1\), start by marking the center at \((-3, 1)\). From this center, measure **5 units** vertically (semi-major axis) and **3 units** horizontally (semi-minor axis) to sketch the ellipse.
The visual helps in better understanding and verifying the solution derived through algebraic manipulation.