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Chapter 12: Problem 27

Sketch the graph of each ellipse. $$ (x-2)^{2}+\frac{(y+1)^{2}}{36}=1 $$

Short Answer

Expert verified
Center at (2, -1), major axis length 12 along y-axis, minor axis length 2 along x-axis.

Step by step solution

01

- Identify the Center

The given ellipse equation is \((x-2)^{2}+\frac{(y+1)^{2}}{36}=1\). The center of an ellipse in standard form \((x-h)^{2}+\frac{(y-k)^{2}}{b^2}=1\) is \( (h, k) \). Here, the center is \((2, -1)\).
02

- Determine the Length of the Major and Minor Axes

In the equation \((x-2)^{2}+\frac{(y+1)^{2}}{36}=1\), compare it with the standard form \( (x-h)^{2}/a^2 + (y-k)^{2}/b^2 = 1 \). Here, the term \( \frac{(y+1)^{2}}{36} \) tells us \((b^2 = 36)\), so \( b = 6 \). Since there's no denominator under \( (x-2)^{2} \, a^2 = 1\), so \( a = 1 \). The major axis length is \( 2b = 2 \times 6 = 12\). The minor axis length is \( 2a = 2 \times 1 = 2\).
03

- Sketch the Ellipse

Start by plotting the center of the ellipse at \( (2, -1) \). From the center, move 6 units up and 6 units down to mark the ends of the major axis (along the y-axis). Then, move 1 unit left and 1 unit right to mark the ends of the minor axis (along the x-axis). Draw a smooth curve to connect these points forming the ellipse.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ellipse equation
An ellipse is a geometric figure that looks like a stretched circle. The general equation for an ellipse with a horizontal major axis is \[\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\]. This equation tells us about the shape and position of the ellipse.
The values of 'h' and 'k' are the coordinates of the center of the ellipse. The terms 'a' and 'b' define the lengths of the stretches along the x-axis and y-axis, respectively. If 'a' and 'b' are equal, the ellipse is a circle.
In the given exercise, the equation is written differently but it follows the same principle.
major axis
The major axis of an ellipse is the longest diameter that runs through the center. It spans from one edge of the ellipse to the other, passing through the center.
In the given equation, \((x-2)^{2} + \frac{(y+1)^{2}}{36} = 1\), we identify 'b' as 6, meaning the major axis is 12 units long because \[2b = 2 \times 6 = 12\]. Here, 'b' corresponds to the length along the y-axis.
This axis is perpendicular to the minor axis and gives the ellipse its elongated shape.
minor axis
The minor axis is the shorter diameter of the ellipse, also running through the center and perpendicular to the major axis.
From the given ellipse equation, \((x-2)^{2} + \frac{(y+1)^{2}}{36} = 1\), we can see that since there is no divisor under the \((x-2)^{2}\) term, we assume 'a = 1'. Therefore, the minor axis is 2 units long because \[2a = 2 \times 1 = 2\].
Even though it鈥檚 shorter, the minor axis is essential for determining the overall shape of the ellipse.
center of ellipse
The center of an ellipse is the point where the major and minor axes intersect. It is denoted by coordinates (h, k) in the standard ellipse equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\].
For the problem at hand, \((x-2)^2 + \frac{(y+1)^2}{36} = 1\), the center is clearly (2, -1).
It鈥檚 crucial to locate the center accurately, as it helps in sketching the ellipse by marking the starting point from which we measure the stretches along the major and minor axes.

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Most popular questions from this chapter

Determine whether the graph of each equation is a circle, parabola, ellipse, or hyperbola. $$ y=x^{2}+1 $$

Sketch the graph of each parabola. $$ y=-\frac{1}{2}(x+1)^{2}+5 $$

Sketch the graph of each ellipse. $$ \frac{(x-3)^{2}}{4}+\frac{(y-1)^{2}}{9}=1 $$

Graph \(y_{1}=\sqrt{x^{2}-1}, y_{2}=-\sqrt{x^{2}-1}, y_{3}=x,\) and \(y_{4}=-x\) to get the graph of the hyperbola \(x^{2}-y^{2}=1\) along with its asymptotes. Use the viewing window \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3 .\) Notice how the branches of the hyperbola approach the asymptotes.

Graph both equations of each system on the same coordinate axes. Solve the system by elimination of variables to find all points of intersection of the graphs. $$\begin{aligned}&(x+1)^{2}+(y-4)^{2}=17\\\&y=x+2\end{aligned}$$
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