Explanation Given information: Given that a > b > 0 and c = a 2 + b 2 , P ( x , y ) is a point on the hyperbola ( x a ) 2 − ( y b ) 2 = 1 . Also F = ( c , 0 ) , e = c a and vertical directrix D at x = a e . Proof: The distance between the point P and F is given by: P F = ( x − c ) 2 + ( y − 0 ) 2 P F = ( x − c ) 2 + y 2 The distance between the point P and the directrix D is given by: P D = | x − a e | Consider the equation P F = e P D . Plugging the values to check if the equation holds true: ( x − c ) 2 + y 2 = e | x − a e | ( ( x − c ) 2 + y 2 ) 2 = ( e | x − a e | ) 2 ( x − c ) 2 + y 2 = e 2 ( x − a e ) 2 ( x − c ) 2 − y 2 = ( c a ) 2 ( x − a ( c a ) ) 2 ( x − c ) 2 + y 2 = c 2 a 2 ( x − a 2 c ) 2 a 2 ( x 2 − 2 c x +

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Chapter 11.5, Problem 66E

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To prove that a point P=(x,y) on the given hyperbola satisfies PF=ePD for the given conditions.

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Chapter 11.5, Problem 65E

Chapter 11.5, Problem 67E

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Chapter 11 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 6PQ Ch. 11.1 - Prob. 7PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E

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To prove that a point P = ( x , y ) on the given hyperbola satisfies P F = e P D for the given conditions.

Solution Summary: The author proves that a point P=(x,y) satisfies PF=ePD for the given conditions. The distance between the point P and the directrix

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To prove that a point P=(x,y) on the given hyperbola satisfies PF=ePD for the given conditions.

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Chapter 11 Solutions