Chapter 10.4, Problem 81AYU

The eccentricity $e$ of a hyperbola is defined as the number $\frac{c}{a}$, where $a$ is the distance of a vertex from the center and $c$ is the distance of a focus from the center. Because $c>a$, it follows that $e>1$. Describe the general shape of a hyperbola whose eccentricity is close to 1. What is the shape if $e$ is very large?

To determine

To explain:

a. The general shape of hyperbola whose eccentricity is close to 1.

Answer to Problem 81AYU

Solution:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

a. If the eccentricity is close to 1 then $c\approx a$ and $b\approx 0$. If $b$ is close to 0, then slopes of the asymptotes are close to zero and the hyperbola becomes very narrow.

Let the equation of the hyperbola be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ the opposite is true.

a. If the eccentricity is close to 1 then the slopes of asymptotes are close to 0, hence the hyperbola is very wide.

Explanation of Solution

Given:

The eccentricity $e$ of a hyperbola is defined as the number $\frac{c}{a}$, where $a$ is the distance of a vertex from the center and $c$ is the distance of a focus from the center. Because $c>a$, it follows that $e>1$.

Calculation:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

a. If the eccentricity is close to 1 then $c\approx a$ and $b\approx 0$. If $b$ is close to 0, then slopes of the asymptotes are close to zero and the hyperbola becomes very narrow.

Let the equation of the hyperbola be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ the opposite is true.

a. If the eccentricity is close to 1 then the slopes of asymptotes are close to 0, hence the hyperbola is very wide.

To determine

To explain:

b. Shape of hyperbola if e is very large.

Answer to Problem 81AYU

Solution:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

b. If the eccentricity is very large, then $c$ is much larger than $a$ and $b$ is very large.

The slope of asymptote becomes very large. Hence, the hyperbola becomes very is wide.

Let the equation of the hyperbola be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ the opposite is true.

b. If the eccentricity is close to 0 then the slopes of asymptotes are very large, hence the hyperbola is very narrow.

Explanation of Solution

Given:

The eccentricity $e$ of a hyperbola is defined as the number $\frac{c}{a}$, where $a$ is the distance of a vertex from the center and $c$ is the distance of a focus from the center. Because $c>a$, it follows that $e>1$.

Calculation:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

b. If the eccentricity is very large, then $c$ is much larger than $a$ and $b$ is very large.

The slope of asymptote becomes very large. Hence, the hyperbola becomes very is wide.

Let the equation of the hyperbola be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ the opposite is true.

b. If the eccentricity is close to 0 then the slopes of asymptotes are very large, hence the hyperbola is very narrow.