Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
1st Edition
ISBN: 9780078682278
Author: McGraw-Hill, Berchie Holliday
Publisher: Glencoe/McGraw-Hill
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Chapter 10.7, Problem 47E

(a)

To determine

To find:the whether the reflector in the telescope is elliptical, parabolic or hyperbolic.

(a)

Expert Solution
Check Mark

Answer to Problem 47E

This conic is an ellipse.

Explanation of Solution

Given:

  9x2−23xy+11y2−240.

Concept used:

The general equation for conic section is:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.

Where A,B,C,D,E and F are constant.

As by changing the values of some of the constants, the shape of the corresponding conic also changes.

If B2−4AC , is less than zero, if a conic exists, it will be either a circle or an ellipse.

If B2−4AC ,equals to zero, if a conic exists, it will be parabola.

If B2−4AC id greater than zero, if a conic exists, it will be hyperbola.

Calculation:

Consider the equation as:

  9x2−23xy+11y2−240.

To determine the minimum angle of rotation needed to transform the graph of this equation to a graph whose axes are on the x and y−axis.

Write the above equation in the standard conic form:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.9x2−23xy+11y2−240.

The angle od rotation is given by tan2θ=BA−C .

So, the angle of rotation θ for the 9x2−23xy+11y2−240 :

  tan2θ=−239−11.θ=12tan−1(−3).θ=−300.

Hence, the angle of rotation θ through which the mirror has been rotated is −300 .

(b)

To determine

To sketch:the graph of the equation using the graphing calculator.

(b)

Expert Solution
Check Mark

Answer to Problem 47E

The graph is ellipse.

Explanation of Solution

Given:

  31x2−103xy+21y2−144=0 .

Concept used:

The general equation for conic section is:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.

Where A,B,C,D,E and F are constant.

As by changing the values of some of the constants, the shape of the corresponding conic also changes.

If B2−4AC , is less than zero, if a conic exists, it will be either a circle or an ellipse.

If B2−4AC ,equals to zero, if a conic exists, it will be parabola.

If B2−4AC id greater than zero, if a conic exists, it will be hyperbola.

Calculation:

Use the angle θ=−300 ,

Find the new equation of the graph.

To find the expressions, replace x and y .

Replace x with x′cos(−300)+y′sin(−300) or x′32−y′12.

And y with −x′sin(−300)+y′cos(−300) or x′12+y′32.

(c)

To determine

To find:the angle through which the mirror has been rotated.

(c)

Expert Solution
Check Mark

Answer to Problem 47E

The angle of rotation θ through which the mirror has been rotated is −300 .

Explanation of Solution

Given:

  9x2−23xy+11y2−24=0.

Concept used:

The general equation for conic section is:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.

Where A,B,C,D,E and F are constant.

  tan2θ=BA−C .

Where θ is the angle.

Calculation:

Substitute the value of x and y in 9x2−23xy+11y2−24=0.

  9(x′32−y′12)2−23(x′32−y′12)(x′12−y′32)+11(x′12−y′32)2−24=0.9(34x′+214y′2−x′y′32)−23(34x′−234y′2+3x′y′4−x′y′4)+11(14x′+234y′2+x′y′32)=0.

It can be further solved as:

  (274x′2+114x′2−32x′2)+(94y′2+64y′2+334y′2)+(1132x′y′−932x′y′−231x′y′)−24=0.

Grouping the like term as:

  8x′2+12y′2=24.x′23+y′22=1.

Hence, therefore the new equation of the graph is x′23+y′22=1 by dividing with 24 throughout.

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Chapter 10.7, Problem 46E
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Chapter 10.7, Problem 48E

Chapter 10 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

Ch. 10.1 - Prob. 1CFUCh. 10.1 - Prob. 2CFUCh. 10.1 - Prob. 3CFUCh. 10.1 - Prob. 4CFUCh. 10.1 - Prob. 5CFUCh. 10.1 - Prob. 6CFUCh. 10.1 - Prob. 7CFUCh. 10.1 - Prob. 8CFUCh. 10.1 - Prob. 9CFUCh. 10.1 - Prob. 10CFU
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