Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. $$ x^{2}+4 y^{2}-2 x+16 y+1=0 $$

Step-by-Step Solution

Answer

The graph is an ellipse centered at (1, -2) with a semi-major axis 4 along the x-axis and semi-minor axis 2 along the y-axis.

Step 1: Rewrite the Equation

The given equation is \( x^2 + 4y^2 - 2x + 16y + 1 = 0 \). Start by rewriting it to group the \( x \) and \( y \) terms together: \[ x^2 - 2x + 4(y^2 + 4y) + 1 = 0 \]

Step 2: Complete the Square for x

Focus on the \( x \) terms \( x^2 - 2x \). To complete the square, add and subtract \( (-1)^2 \). This results in:\[ (x-1)^2 - 1 \] Replace \( x^2 - 2x \) with this expression in the equation.

Step 3: Complete the Square for y

For the \( y \) terms, look at \( y^2 + 4y \). To complete the square, you add and subtract \( "+4" \) (which is \( (4/2)^2 \)). Thus,\[ y^2 + 4y = (y+2)^2 - 4 \]Substitute this back into the main equation.

Step 4: Substitute and Simplify the Equation

Substitute the completed squares back into the equation:\[ (x-1)^2 - 1 + 4((y+2)^2 - 4) + 1 = 0 \]This simplifies to:\[ (x-1)^2 + 4(y+2)^2 = 16 \]

Step 5: Identify the Conic Section

The equation \( (x-1)^2 + 4(y+2)^2 = 16 \) is in the form of a standard ellipse equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Identifying:- Center \((h, k) = (1, -2)\)- \( a^2 = 16 \), so \( a = 4 \)- \( b^2 = 4 \), so \( b = 2 \)Thus, this is an ellipse centered at \((1, -2)\) with semi-major axis 4 (along x-axis) and semi-minor axis 2 (along y-axis).

Step 6: Sketch the Graph

To sketch the graph of the ellipse:1. Plot the center of the ellipse at \((1, -2)\).2. Draw the major axis along the x-axis, 4 units right and left of the center.3. Draw the minor axis along the y-axis, 2 units up and down from the center.4. Sketch the ellipse shape fitting these dimensions.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This helps in simplifying equations, especially when dealing with conic sections like ellipses.

To complete the square for an expression like \(x^2 - 2x\), follow these steps:

• Take the coefficient of \(x\), which is \(-2\), and divide it by 2, giving you \(-1\).
• Square the result, so \((-1)^2 = 1\).
• Add and subtract this square inside the expression: \(x^2 - 2x + 1 - 1 = (x-1)^2 - 1\).

Similarly, for \(y^2 + 4y\), you:

• Take \(4\), divide by 2 to get \(2\), then square it to obtain \(4\).
• Add and subtract this from the equation: \(y^2 + 4y + 4 - 4 = (y+2)^2 - 4\).

Completing the square effectively rewrites the quadratic terms into squared terms, making it easier to identify and work with the equation of a conic section.

Conic Sections

Conic sections are curves generated by the intersection of a plane with a cone. The four types of conic sections are circles, ellipses, parabolas, and hyperbolas, each with distinct equations and properties.

Ellipses arise when the plane intersects the cone at an angle, causing a closed, oval-shaped curve. These shapes follow a standard form equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where:

• \((h, k)\) is the center of the ellipse.
• \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively.
• If \(a > b\), the ellipse is stretched along the x-axis, and if \(b > a\), it is stretched along the y-axis.

Understanding conic sections through their standard form simplifies the process of analyzing and sketching these curves on a graph.

Ellipse Equation

The ellipse equation is an essential element in describing the shape and features of an ellipse. The equation is typically expressed as \((x-1)^2 + 4(y+2)^2 = 16\).

To rewrite this ellipse equation in standard form, divide each term by 16:

• \(\frac{(x-1)^2}{16} + \frac{4(y+2)^2}{16} = 1\).
• Simplifying, this becomes \(\frac{(x-1)^2}{16} + \frac{(y+2)^2}{4} = 1\).

From this equation, it's easy to identify the ellipse's characteristics:

• Center at \((1, -2)\).
• \(a^2 = 16\), functioning as the semi-major axis \(a = 4\), aligned along the x-axis.
• \(b^2 = 4\), functioning as the semi-minor axis \(b = 2\), aligned along the y-axis.

The equation provides a precise description of how the ellipse is oriented and scaled within the coordinate plane.

Graph Sketching

Graph sketching involves plotting the visual representation of an equation on a coordinate plane. For an ellipse, it means identifying key components and then sketching them accordingly.

Begin sketching the graph of an ellipse with these steps:

• Determine and plot the center, which in this case is \((1, -2)\).
• From the center, move \(a = 4\) units along the x-axis for the vertices of the major axis, placing them at \((-3, -2)\) and \((5, -2)\).
• Do the same for the y-axis using \(b = 2\), placing edges at \((1, 0)\) and \((1, -4)\).
• Connect these points in a smooth, oval shape to produce the complete sketch of the ellipse.

Graph sketching is a vital skill that transforms equations into visual understanding, clearly illustrating the size, orientation, and position of ellipses and other conic sections.