Explanation Given: P F 1 2 − P F 2 2 = 4 x 0 c and P F 1 + P F 2 = 2 a The relation e = c a . A x + B y = 1 where A = x 0 a 2 and B = y 0 b 2 Calculation: a. The equation of the tangent line at the point P is A x + B y = 1 From the above graph, we can see that: P F 1 2 = ( x 0 + c ) 2 + y 0 2 and P F 2 2 = ( x 0 − c ) 2 + y 0 2 Now, P F 1 2 − P F 2 2 = ( x 0 + c ) 2 + y 0 2 − ( x 0 − c ) 2 − y 0 2 = ( x 0 + c ) 2 − ( x 0 − c ) 2 = 4 c x 0 ( Put c=ae ) P F 1 2 − P F 2 2 = 4 a e x 0 ... ( 1 ) It is given that P F 1 + P F 2 = 2 a ... ( 2 ) So, from the equation (1) and equation (2) we get: P F 1 − P F 2 = 2 e x 0 ... ( 3 ) Now, add the equation (2) and equation (3) and we get: 2 P F 1 = 2 a + 2 e x 0 P F 1 = a + e x 0 Now, subtract equation (2) and equation (3) and we get: 2 P F 2 = 2 a − 2 e x 0 P F 2 = a − e x 0 Thus, it is proved that P F 1 = a + x 0 e and P F 2 = a − x 0 e . b...

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 12.5, Problem 72E

To determine

a. That $P F_{1} = a + x_{0} e and P F_{2} = a - x_{0} e$

b. That $\frac{F_{1} R_{1}}{P F_{1}} = \frac{F_{2} R_{2}}{P F_{2}}$

c. That $\sin θ_{1} = \sin θ_{_{2}}$ and conclude that $θ_{1} = θ_{_{2}}$

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Chapter 12.5, Problem 71E

Chapter 12.5, Problem 73E

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Chapter 12 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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

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a. That P F 1 = a + x 0 e and P F 2 = a − x 0 e b. That F 1 R 1 P F 1 = F 2 R 2 P F 2 c. That sin θ 1 = sin θ 2 and conclude that θ 1 = θ 2

Solution Summary: The author calculates that P F 1 = a + x 0 e and P

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a. That $P F_{1} = a + x_{0} e and P F_{2} = a - x_{0} e$

b. That $\frac{F_{1} R_{1}}{P F_{1}} = \frac{F_{2} R_{2}}{P F_{2}}$

c. That $\sin θ_{1} = \sin θ_{_{2}}$ and conclude that $θ_{1} = θ_{_{2}}$

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Chapter 12 Solutions

a. That P F 1 = a + x 0 e and P F 2 = a − x 0 e b. That F 1 R 1 P F 1 = F 2 R 2 P F 2 c. That sin θ 1 = sin θ 2 and conclude that θ 1 = θ 2

Solution Summary: The author calculates that P F 1 = a + x 0 e and P

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a. That PF1=a+x0e and PF2=a−x0e b. That F1R1PF1=F2R2PF2 c. That sinθ1=sinθ2 and conclude that θ1=θ2

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Chapter 12 Solutions

a. That $P F_{1} = a + x_{0} e and P F_{2} = a - x_{0} e$

b. That $\frac{F_{1} R_{1}}{P F_{1}} = \frac{F_{2} R_{2}}{P F_{2}}$

c. That $\sin θ_{1} = \sin θ_{_{2}}$ and conclude that $θ_{1} = θ_{_{2}}$