Chapter 10: Q35. (page 677)

In Problems 31– 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 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 $60^{\circ} .$

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

$θ$ need to be acute.

So, $c o t 2 θ = - \frac{\sqrt{3}}{3} 2 θ = 120 θ = 60^{\circ}$

Step 3. Finding the values of x,y

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

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

Step 4. Substitute these values in the original equation

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

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

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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In Problems 31– 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 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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In Problems 31– 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 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

Applied Mathematics

Theoretical and Mathematical Physics

Calculus

Logic and Functions

Pure Maths

Geometry

Study anywhere. Anytime. Across all devices.

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