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Chapter 10: Problem 33

Ellipses Centered at \((h, k)\) Graph. $$4(x+3)^{2}+4(y+1)^{2}-10=90$$

Short Answer

Expert verified
Center: \((-3, -1)\), Axes lengths: 5 units horizontally and vertically.

Step by step solution

01

Simplify the Equation

Begin by simplifying the given equation: \[4(x+3)^{2}+4(y+1)^{2}-10=90\] Add 10 to both sides to isolate the terms involving \(x\) and \(y\). \[4(x+3)^{2}+4(y+1)^{2}=100\]
02

Divide Both Sides by 100

Divide the entire equation by 100 to get it in the standard form of an ellipse equation. \[ \frac{4(x+3)^2}{100} + \frac{4(y+1)^2}{100} = \frac{100}{100} \] This simplifies to: \[ \frac{(x+3)^2}{25} + \frac{(y+1)^2}{25} = 1 \]
03

Identify the Center and Axes Length

The standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) represents an ellipse centered at \((h, k)\). Identify the center \((h, k)\) from the equation: Center: \((-3, -1)\). Here, both denominators are 25 so \(a^2 = b^2 = 25\), thus, \(a = b = 5\). This indicates the lengths of the major and minor axes.
04

Graph the Ellipse

Plot the center of the ellipse at \((-3, -1)\). Since both axes lengths are the same (5 units), plot points 5 units away from the center both horizontally and vertically: - Right: \((-3 + 5, -1) = (2, -1)\)- Left: \((-3 - 5, -1) = (-8, -1)\)- Upwards: \((-3, -1 + 5) = (-3, 4)\)- Downwards: \((-3, -1 - 5) = (-3, -6)\)

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Key Concepts

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

Standard Form of an Ellipse
Before diving into graphing an ellipse, it's important to understand its standard form. The standard form of an ellipse is represented by the equation:
\(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
Here, \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis (half the length of the major axis), and \(b\) is the semi-minor axis (half the length of the minor axis). This form helps in easily identifying key attributes of the ellipse, like its center and axes lengths. In our example:
\(\frac{(x+3)^2}{25} + \frac{(y+1)^2}{25} = 1\).
We see that both denominators are equal, indicating a circular shape where the semi-major and semi-minor axes are the same length.
Identifying the Center of an Ellipse
Identifying the center of an ellipse is straightforward once it's in standard form. You recognize the terms \(x - h\) and \(y - k\) in the equation. They effectively shift the ellipse horizontally and vertically. In our example:
\(\frac{(x+3)^2}{25} + \frac{(y+1)^2}{25} = 1\),
we have \((x + 3)\) and \((y + 1)\).
This tells us that the center of the ellipse is at \((h, k) = (-3, -1)\).
Recognizing this allows for accurate graphing and understanding of the ellipse's position in the coordinate plane.
Graphing Ellipses
Once you've identified the center, you can start graphing the ellipse. Plot the center first, then use the lengths of the semi-major and semi-minor axes to draw it. In our case, the center is at \((-3, -1)\).
The next step is to use the values of \(a\) and \(b\), which are both 5 in our example, to plot key points:
  • Right: \((-3 + 5, -1) = (2, -1)\)
  • Left: \((-3 - 5, -1) = (-8, -1)\)
  • Upwards: \((-3, -1 + 5) = (-3, 4)\)
  • Downwards: \((-3, -1 - 5) = (-3, -6)\)
Connect these points smoothly to complete the ellipse. Make sure to maintain the shape as dictated by the lengths of the axes. This forms the complete graph of the ellipse.

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Most popular questions from this chapter

Solve. Remember that graphs can be used to confirm all real solutions. $$\begin{aligned}&a b-b^{2}=-4\\\&a b-2 b^{2}=-6\end{aligned}$$

Solve. Remember that graphs can be used to confirm all real solutions. $$\begin{aligned}&x^{2}+y^{2}=16\\\&y^{2}-2 x^{2}=10\end{aligned}$$

Solve. The product of two numbers is 90. The sum of their squares is 261. Find the numbers.

Solve. Remember that graphs can be used to confirm all real solutions. $$\begin{aligned}&3 x y+x^{2}=34\\\&2 x y-3 x^{2}=8\end{aligned}$$

Find an equation of an ellipse that contains the following points. $$(-7,0),(7,0),(0,-10), \text { and }(0,10)$$
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