Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Question

Chapter 10, Problem 10CT

To determine

The analysis of the equation $41x2−24xy+34y2−25=0$ by rotating the axis, so that there is no $xy−$ term. Graph the new equation.

Expert Solution & Answer

Answer to Problem 10CT

Solution:

The equation is $x'21+y'212=1$ .

The graph of the equation $x'21+y'212=1$ is

Precalculus Enhanced with Graphing Utilities, Chapter 10, Problem 10CT , additional homework tip 1

Explanation of Solution

Given information:

The equation $41x2−24xy+34y2−25=0$

To transform the equation of the form $Ax2+Bxy+Cy2++Dx+Ey+F=0, B≠0$ into an equation in $x'$ and $y'$ without $x'y'−$ term, rotate the axis through an angle $θ$ that satisfies the equation $cot(2θ)=A−CB$ .

Compare the equation $41x2−24xy+34y2−25=0$ with $Ax2+Bxy+Cy2++Dx+Ey+F=0, B≠0$ gives $A=41,B=−24$ and $C=34$ ,

$⇒cot(2θ)=41−34−24$

$⇒cot(2θ)=−724$

Since $cot(2θ)<0$ , then $90∘<2θ<180∘$

$⇒45∘<θ<90∘$ .

Precalculus Enhanced with Graphing Utilities, Chapter 10, Problem 10CT , additional homework tip 2

$cot(2θ)=−724$ it follows that $cos(2θ)=−725$

$sinθ=1−cos(2θ)2=1−(−725)2=1625=45$

$cosθ=1+cos(2θ)2=1+(−725)2=925=35$

$θ=cos−1(35)≈53.13°$

Now, $x=x'cosθ−y'sinθ$ and $y=x'sinθ+y'cosθ$

$⇒x=35x'−45y'$ and $y=45x'+35y'$

Substituting these values in the equation $41x2−24xy+34y2−25=0$ ,

$⇒41(35x'−45y')2−24(35x'−45y')(45x'+35y')+34(45x'+35y')2−25=0$

By simplifying,

$⇒41(3x'−4y')225−24(3x'−4y')(4x'+3y')25+34(4x'+3y')225=25$

Multiplying by $25$ to both sides,

$⇒41(3x'−4y')2−24(3x'−4y')(4x'+3y')+34(4x'+3y')2=625$

$⇒41(9x'2−24x'y'+16y'2)−24(12x'2−7x'y'−12y'2)+34(16x'2+24x'y'+9y'2)=625$

$⇒369x'2−574x'y'+656y'2−288x'2+169x'y'+288y'2+544x'2+816x'y'+306y'2=625$

$⇒625x'2+1250y'2=625$

Dividing by $625$ on both sides of equation,

$⇒x'2+2y'2=1$

$⇒x'21+y'212=1$

This is an equation of an ellipse with center at $(h,k)=(0,0)$ in the $x'y'−$ plane.

Thus, $a=1=1$ , $b=12=22$ and the vertices are $(h+a,k)=(1,0)$ , $(h−a,k)=(−1,0)$ in the $x'y'−$ plane.

$c2=a2−b2=1−12=12$

$⇒c=22$

The foci are at $(h±c,k)=(±22,0)$ in the $x'y'−$ plane.

The minor axis intercept are $(h,k±b)=(0,±22)$

The graph of the equation $x'21+y'212=1$ is as follow:

Precalculus Enhanced with Graphing Utilities, Chapter 10, Problem 10CT , additional homework tip 3

Chapter 10, Problem 9CT

Chapter 10, Problem 11CT

Chapter 10 Solutions

Precalculus Enhanced with Graphing Utilities

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Precalculus

Knowledge Booster

The analysis of the equation 41 x 2 − 24 x y + 34 y 2 − 25 = 0 by rotating the axis, so that there is no x y − term. Graph the new equation.

Answer to Problem 10CT

Explanation of Solution

Chapter 10 Solutions

Additional Math Textbook Solutions