Chapter 9: Q. 59 (page 773)

Prove that for an ellipse or a hyperbola the eccentricity is given by
$e = \frac{the distance between the foci}{the distance between the vertices}$

Short Answer

Expert verified

Hence proved.

Step by step solution

Step 1. Given information.

We are given,

$e = \frac{the distance between the foci}{the distance between the vertices}$

Step 2. Explanation.

Now,

$Let the foci of the ellipse are (\pm \sqrt{A^{2} - B^{2}}, 0) . Formula for the distance = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} Distance = \sqrt{{(\sqrt{A^{2} - B^{2}} - (- \sqrt{A^{2} - B^{2}}))}^{2} + (0 - 0)^{2}} Distance = \sqrt{{(2 \sqrt{A^{2} - B^{2}})}^{2}} Thus the distance between the foci is, Distance = 2 \sqrt{A^{2} - B^{2}} W e h a v e t h e e c c e n t r i c i t y e = \frac{\sqrt{A^{2} - B^{2}}}{A} Then e = \frac{2 \sqrt{A^{2} - B^{2}}}{2 A} Thus, e = \frac{the distance between foci}{the distance between the vertices}$

Hence proved.

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Prove that for an ellipse or a hyperbola the eccentricity is given by
$e = \frac{the distance between the foci}{the distance between the vertices}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

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Prove that for an ellipse or a hyperbola the eccentricity is given bye=the distance between the focithe distance between the vertices

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Applied Mathematics

Mechanics Maths

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Statistics

Study anywhere. Anytime. Across all devices.

Prove that for an ellipse or a hyperbola the eccentricity is given by
$e = \frac{the distance between the foci}{the distance between the vertices}$