Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts.. $$ 4(x-5)^{2}+9(y+2)^{2}=36 $$

Step-by-Step Solution

Answer

The ellipse is centered at \((5, -2)\) with intercepts at \((8, -2), (2, -2), (5, 0), (5, -4)\).

Step 1: Identify the Conic Section

Recognize that the given equation \(4(x-5)^2 + 9(y+2)^2 = 36\) is the equation of an ellipse. This can be seen because it is in the form \(Ax^2 + By^2 = C\), where both \(A\) and \(B\) are positive, indicating an ellipse.

Step 2: Put the Equation in Standard Form

To convert the equation into the standard form of an ellipse, divide the entire equation by 36:\[\frac{4(x-5)^2}{36} + \frac{9(y+2)^2}{36} = 1\]Simplifying gives:\[\frac{(x-5)^2}{9} + \frac{(y+2)^2}{4} = 1\]This is now in the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).

Step 3: Determine Center, Semi-Axes, and Orientation

From the standard form \(\frac{(x-5)^2}{9} + \frac{(y+2)^2}{4} = 1\), identify:- Center \((h, k) = (5, -2)\)- \(a^2 = 9\), so \(a = 3\) (horizontal axis)- \(b^2 = 4\), so \(b = 2\) (vertical axis)The ellipse is horizontally oriented, as \(a > b\).

Step 4: Find the X-Intercepts

Set \(y + 2 = 0\) (plug \(y = -2\) into the ellipse equation):\[\frac{(x-5)^2}{9} = 1\]This gives \((x-5)^2 = 9\), so \(x - 5 = \pm 3\). Therefore, \(x = 8\) and \(x = 2\). The \(x\)-intercepts are \((8, -2)\) and \((2, -2)\).

Step 5: Find the Y-Intercepts

Set \(x - 5 = 0\) (plug \(x = 5\) into the ellipse equation):\[\frac{9(y+2)^2}{36} = 1\]This gives \((y+2)^2 = 4\), so \(y + 2 = \pm 2\). Therefore, \(y = 0\) and \(y = -4\). The \(y\)-intercepts are \((5, 0)\) and \((5, -4)\).

Step 6: Check for Symmetries

Due to the form of the equation, the ellipse is symmetric about the horizontal line \(y = -2\), the vertical line \(x = 5\), and also symmetric about its center \((5, -2)\).

Step 7: Plot the Ellipse

Draw a coordinate grid and plot the center at \((5, -2)\). Since \(a = 3\), draw the ellipse extending 3 units left and right from the center horizontally to \((8, -2)\) and \((2, -2)\). Since \(b = 2\), draw the ellipse extending 2 units up and down from the center vertically to \((5, 0)\) and \((5, -4)\). Connect these points forming a smooth, oval shape.

Key Concepts

Conic Sections

Conic sections are the curves that can be formed by the intersection of a plane and a double-napped cone. There are four basic types of conic sections: circles, ellipses, parabolas, and hyperbolas. Each of these is distinguishable based on the plane's angle relative to the cone. For example, an ellipse is formed when the plane cuts through the cone at an angle, but does not cut through the base. It resembles an elongated circle. In this exercise, we recognize that the equation given \[4(x-5)^2+9(y+2)^2=36\] defined an ellipse due to the form \[Ax^2 + By^2 = C\] where both coefficients \(A\) and \(B\) are positive units.

Ellipse Graph

Graphing an ellipse involves converting its equation to a standard form so that key characteristics can be identified easily. This involves ensuring the constant on the right side of the equation is 1. For the given problem, dividing the whole equation by 36 gives us the standard form: \[\frac{(x-5)^2}{9} + \frac{(y+2)^2}{4} = 1\] This allows us to identify the ellipse's center, which in this case is at the point \((5, -2)\). The values \(a^2=9\) and \(b^2=4\) indicate the lengths of the semi-axes: \(a=3\) for the horizontal semi-axis and \(b=2\) for the vertical. Hence, the ellipse is oriented horizontally, spanning 3 units on either side from the center Understanding these dimensions is crucial, as they help in sketching the ellipse accurately. To draw the graph, remember:

• Start at the center point \((5, -2)\)
• Move horizontally 3 units left to \((2, -2)\) li>Move 3 units right to \((8, -2)\), completing the horizontal span.
• Move vertically 2 units up and down from the center to points \((5, 0)\) and \((5, -4)\)

Symmetries in Ellipse

Ellipses exhibit symmetry, which simplifies their analysis on the graph. This particular ellipse is symmetric about both its axes and its center due to its bisected nature through the axes and origin. Symmetry is visible when reflection or rotation doesn’t change the ellipse's appearance. For this exercise, checking its symmetry reveals:

• Horizontal symmetry about line \(y=-2\)
• Vertical symmetry about line \(x=5\)
• Central symmetry about its center \((5, -2)\)

This means any part of the ellipse you examine will have a corresponding part on the opposite side, which greatly aids when graphing by ensuring accuracy and coherence.

Finding Intercepts

To find the intercepts of an ellipse, set either \(x\) or \(y\) to zero in the standard form and solve for the other variable. These intercepts are the points where the ellipse meets the axes. For the given equation \[\frac{(x-5)^2}{9} + \frac{(y+2)^2}{4} = 1,\]Finding the X-intercepts:

• Set \(y + 2 = 0\) (giving \(y = -2\)), leading to \[\frac{(x-5)^2}{9} = 1\]
• Solve this for \(x\): \((x-5)^2 = 9\), thus \(x-5=\pm 3\).
• Yields X-intercepts at \((8, -2)\) and \((2, -2)\)

Finding the Y-intercepts:

• Set \(x - 5 = 0\) (giving \(x = 5\)), leading to \[\frac{9(y+2)^2}{36} = 1\]
• Solve this for \(y\): \((y + 2)^2 = 4\), thus \(y + 2 =\pm 2\).
• Gives Y-intercepts at \((5, 0)\) and \((5, -4)\)

Thus, these intercept points guide the graphing and analysis of the ellipse on a coordinate plane.