Q35. In Problems 31鈥� 42, rotate the... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 10: Q35. (page 677)

In Problems 31鈥� 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21鈥�30 for Problems 31鈥� 40. $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$

Short Answer

Expert verified

The ellipse is rotated through an angle of $60^{鈭�} .$

x = \frac{1}{2} x' - \frac{\sqrt{3}}{2} y' y = \frac{\sqrt{3}}{2} x' + \frac{1}{2} y'

Step by step solution

Step 1. Given Information

The given equation is $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$ .

We need to determine the appropriate rotation formula so that new equation does not contain $x y$ term.

Step 2. Finding acute angle

Here $A = 13, B = - 6 \sqrt{3}, C = 7$

role="math" localid="1646931475071"

c o t 2 胃 = \frac{A - C}{B} = \frac{13 - 7}{- 6 \sqrt{3}} c o t 2 胃 = \frac{6}{- 6 \sqrt{3}} = - \frac{\sqrt{3}}{3}

$胃$ need to be acute.

So, $c o t 2 胃 = - \frac{\sqrt{3}}{3} 2 胃 = 120 胃 = 60^{鈭�}$

Step 3. Finding the values of x,y

$x = x' \cos 60^{鈭�} - y' \sin 60^{鈭�} x = \frac{1}{2} x' - \frac{\sqrt{3}}{2} y'$

Similarly, $y = \frac{\sqrt{3}}{2} x' + \frac{1}{2} y'$

Step 4. Substitute these values in the original equation

$13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$

$\frac{13}{4} {(x' - \sqrt{3} y')}^{2} - \frac{6 \sqrt{3}}{2 脳 2} (x' - \sqrt{3} y') (\sqrt{3} x' + y') + \frac{7}{4} {(\sqrt{3} x' + y')}^{2} - 16 = 0 \frac{13}{4} (x'^{2} - 2 \sqrt{3} x' y' + 3 y'^{2}) - \frac{3 \sqrt{3}}{2} (\sqrt{3} x'^{2} + x' y' - 3 x' y' + \sqrt{3} y'^{2}) + \frac{7}{4} (3 x'^{2} + 2 \sqrt{3} x' y' + y'^{2}) = 16$

Step 5. Simplify the above Expression

$\frac{13}{4} x'^{2} + \frac{39}{4} y'^{2} - \frac{13 \sqrt{3}}{2} x' y' - \frac{9}{2} x'^{2} - \frac{3 \sqrt{3}}{2} x' y' + \frac{9 \sqrt{3}}{2} x' y' - \frac{9}{2} y'^{2} + \frac{21}{4} x'^{2} + \frac{7}{4} y'^{2} + \frac{7 \sqrt{3}}{2} x' y' - 16 = 0 \frac{13}{4} x'^{2} - \frac{9}{2} x'^{2} + \frac{39}{4} y'^{2} - \frac{9}{2} y'^{2} - \frac{13 \sqrt{3}}{2} x' y' - \frac{3 \sqrt{3}}{2} x' y' + \frac{9 \sqrt{3}}{2} x' y + \frac{7 \sqrt{3}}{2} x' y' = 16 \frac{5}{4} x'^{2} + \frac{21}{4} y'^{2} = 16$

Step 6. Graphing Equation

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91影视

In Problems 31鈥� 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21鈥�30 for Problems 31鈥� 40. $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding acute angle

Step 3. Finding the values of x,y

Step 4. Substitute these values in the original equation

Step 5. Simplify the above Expression

Step 6. Graphing Equation

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91影视

In Problems 31鈥� 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21鈥�30 for Problems 31鈥� 40.13x2-63xy+7y2-16=0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding acute angle

Step 3. Finding the values of x,y

Step 4. Substitute these values in the original equation

Step 5. Simplify the above Expression

Step 6. Graphing Equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Mechanics Maths

Statistics

Probability and Statistics

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Calculus

Study anywhere. Anytime. Across all devices.

In Problems 31鈥� 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21鈥�30 for Problems 31鈥� 40. $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} - 16 = 0$