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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 10: Q34. (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.
$3 x^{2} - 10 x y + 3 y^{2} - 32 = 0$

Short Answer

Expert verified

The hyperbola is rotated through an angle of $45^{鈭�} .$

Rotated equation is $\frac{y'^{2}}{4} - \frac{x'^{2}}{16} = 1$

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

Step by step solution

01

Step 1. Given Information

The given equation is $3 x^{2} - 10 x y + 3 y^{2} - 32 = 0$ .

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

02

Step 2. Finding acute angle

Here $A = 3, B = - 10, C = 3$

$c o t 2 胃 = \frac{A - C}{B} = \frac{3 - 3}{- 10} = 0$

$胃$ need to be acute.

$c o t 2 胃 = 0 2 胃 = 90 胃 = 45^{鈭�}$

03

Step 3. Finding the values of x,y

$x = x' \cos (45^{鈭�}) - y' \sin (45^{鈭�}) x = \frac{1}{\sqrt{2}} x' - \frac{1}{\sqrt{2}} y'$

and $y = \frac{1}{\sqrt{2}} x' + \frac{1}{\sqrt{2}} y'$

04

Step 4. Substitute these values in the original Equation

$3 x^{2} - 10 x y + 3 y^{2} - 32 = 0 \frac{3}{2} {(x' - y')}^{2} - \frac{10}{\sqrt{2} 脳 \sqrt{2}} (x' - y') (x' + y') + \frac{3}{2} {(x' + y')}^{2} - 32 = 0$

05

Step 5. Simplify the above Equation

$\frac{3}{2} (x'^{2} - 2 x' y' + y'^{2}) - 5 (x'^{2} - y'^{2}) + \frac{3}{2} (x'^{2} - 2 x' y' + y'^{2}) - 32 = 0 \frac{3}{2} x'^{2} + \frac{3}{2} x'^{2} - 5 x'^{2} + \frac{3}{2} y'^{2} + \frac{3}{2} y'^{2} + 5 y'^{2} = 32 - 2 x'^{2} + 8 y'^{2} = 32 \frac{y'^{2}}{4} - \frac{x'^{2}}{16} = 1$

This is the equation of hyperbola.

06

Step 6. Graphing Equation

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The point that is symmetric with respect to the x-axis to the point (-2, 5) is ________.

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