Explanation Given: The perpendicular distances F 1 R 1 and F 2 R 2 Calculation: The standard equation of an ellipse will be given by: ( x a ) 2 + ( y b ) 2 = 1 ... ( 1 ) Equation of the tangent to this ellipse at the point ( x 0 , y 0 ) will be given by: ( x x 0 a ) 2 + ( y y 0 b ) 2 = 1 ... ( 2 ) Now, simplify the equation (2) and we get: ( x x 0 a ) 2 + ( y y 0 b ) 2 = 1 b 2 x 2 x 0 2 + a 2 y 2 y 0 2 − a 2 b 2 = 0 ... ( 3 ) So, the length of perpendiculars on this tangent from the two foci is given by: F 1 R 1 = b 2 x 0 ( − c ) − a 2 b 2 b 4 x 0 2 + a 4 y 0 2 F 2 R 2 = b 2 x 0 ( c ) − a 2 b 2 b 4 x 0 2 + a 4 y 0 2 Now, multiply these perpendicular distances and we get: F 1 R 1 × F 2 R 2 = b 2 x 0 ( − c ) − a 2 b 2 b 4 x 0 2 + a 4 y 0 2 × b 2 x 0 ( c ) − a 2 b 2 b 4 x 0 2 + a 4 y 0 2 F 1 R 1 × F 2 R 2 = b 4 ( a 4 − c 2 x 0 2 ) b 4 x 0 2 + a 4 y 0 2 .....

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ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 12, Problem 55CRE

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That the product of the perpendicular distances $F_{1} R_{1}$ and $F_{2} R_{2}$ from the foci to a tangent line of an ellipse is equal to the square $b^{2}$ of the semi minor axes

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Chapter 12, Problem 54CRE

Chapter 13.1, Problem 1PQ

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Chapter 12 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

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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That the product of the perpendicular distances F 1 R 1 and F 2 R 2 from the foci to a tangent line of an ellipse is equal to the square b 2 of the semi minor axes

Solution Summary: The author proves that the product of the perpendicular distances F 1 R 1 and F 2 R 2 from the foci to a tangent line of an ellipse is equal to the square b

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That the product of the perpendicular distances $F_{1} R_{1}$ and $F_{2} R_{2}$ from the foci to a tangent line of an ellipse is equal to the square $b^{2}$ of the semi minor axes

Want to see the full answer?

Chapter 12 Solutions

That the product of the perpendicular distances F 1 R 1 and F 2 R 2 from the foci to a tangent line of an ellipse is equal to the square b 2 of the semi minor axes

Solution Summary: The author proves that the product of the perpendicular distances F 1 R 1 and F 2 R 2 from the foci to a tangent line of an ellipse is equal to the square b

Videos

That the product of the perpendicular distances F1R1 and F2R2 from the foci to a tangent line of an ellipse is equal to the square b2 of the semi minor axes

Want to see the full answer?

Chapter 12 Solutions

That the product of the perpendicular distances $F_{1} R_{1}$ and $F_{2} R_{2}$ from the foci to a tangent line of an ellipse is equal to the square $b^{2}$ of the semi minor axes