a. With A and B that α 1 + c β 1 = α 2 − c β 2 = A B b. By using (a) and the distance formula to show that F 1 R 1 F 2 R 2 = β 1 β 2 c. By using (a) and the equation of the tangent line in Exercise 68 to show that β 1 = B ( 1 + A c ) A 2 + B 2 , β 2 = B ( 1 − A c ) A 2 + B 2

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Answer 1

Question

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

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(#1) Consider the solid bounded below by z = x² and above by z = 4-y². If we were to project this solid down onto the xy-plane, you should be able to use algebra to determine the 2D region R in the xy-plane for the purposes of integration. Which ONE of these limite of integration would correctly describe R? (a) y: x24x: -22 - (b) y: 22 x: 04-y² (c) y: -√√4-x2. →√√4x²x: −2 → 2 (d) z: 24-y² y: -2 → 2 (e) None of the above

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Answer 2

Question

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Expert Solution & Answer

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Answer 3

Question

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Expert Solution & Answer

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Answer 4

Question

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Expert Solution & Answer

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Answer 5

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Answer 6

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Answer 7

Chapter 12.5, Problem 69E

To determine

a. With A and B that $\frac{α_{1} + c}{β_{1}} = \frac{α_{2} - c}{β_{2}} = \frac{A}{B}$

b. By using (a) and the distance formula to show that

$\frac{F_{1} R_{1}}{F_{2} R_{2}} = \frac{β_{1}}{β_{2}}$

c. By using (a) and the equation of the tangent line in Exercise 68 to show that

$β_{1} = \frac{B (1 + A c)}{A^{2} + B^{2}}, β_{2} = \frac{B (1 - A c)}{A^{2} + B^{2}}$

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

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Answer 11

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

a. With A and B that α 1 + c β 1 = α 2 − c β 2 = A B b. By using (a) and the distance formula to show that F 1 R 1 F 2 R 2 = β 1 β 2 c. By using (a) and the equation of the tangent line in Exercise 68 to show that β 1 = B ( 1 + A c ) A 2 + B 2 , β 2 = B ( 1 − A c ) A 2 + B 2

Solution Summary: The author explains that the equation of the tangent line at the point P is A x + B y = 1 and it is given that R 1 = ( 1, 1) and R 2 =

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Chapter 12 Solutions

a. With A and B that α 1 + c β 1 = α 2 − c β 2 = A B b. By using (a) and the distance formula to show that F 1 R 1 F 2 R 2 = β 1 β 2 c. By using (a) and the equation of the tangent line in Exercise 68 to show that β 1 = B ( 1 + A c ) A 2 + B 2 , β 2 = B ( 1 − A c ) A 2 + B 2

Solution Summary: The author explains that the equation of the tangent line at the point P is A x + B y = 1 and it is given that R 1 = ( 1, 1) and R 2 =

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a. With A and B that α1+cβ1=α2−cβ2=AB b. By using (a) and the distance formula to show that F1R1F2R2=β1β2 c. By using (a) and the equation of the tangent line in Exercise 68 to show that β1=B(1+Ac)A2+B2, β2=B(1−Ac)A2+B2

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Chapter 12 Solutions