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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Question

Chapter 10.7, Problem 46E

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 46E

This conic is an ellipse.

Explanation of Solution

Given:

31x2−103xy+21y2=144.

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:

31x2−103xy+21y2=144.

Which models the shape of a reflecting mirror in a telescope.

To determine whether the reflector is elliptical, parabolic, or hyperbolic.

Write the above equation in the standard conic form:

Ax2+Bxy+Cy2+Dx+Ey+F=0.31x2−103xy+21y2−144=0.

Use the discriminant of the equation to identify the conic as:

B2−4AC=(−103)2−4.31.21=−2304.B2−4AC<0 and A≠C.

Hence, this conic is an ellipse.

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 46E

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:

To sketch the graph of the equation:

31x2−103xy+21y2−144=0 .

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition, Chapter 10.7, Problem 46E

Hence, the graph is ellipse.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 46E

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

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.

tan2θ=BA−C .

Where θ is the angle.

Calculation:

For the standard equation of a conic Ax2+Bxy+Cy2+Dx+Ey+F=0. the angle of rotation θ is given by tan2θ=BA−C .

So, the angle of rotation θ for 31x2−103xy+21y2−144=0 is:

tan2θ=−10331−21.θ=12tan−1(−1.73).⇒12(−59.970).⇒−300.

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

Chapter 10.7, Problem 45E

Chapter 10.7, Problem 47E

Chapter 10 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

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