Chapter A5.2, Problem 21E

To determine

To find: the eccentricity and standard-form equation.

Answer to Problem 21E

Eccentricity is $e=\sqrt{2}$

The hyperbola’s standard-form equation is $\frac{x^2}{8}-\frac{y^2}{8}=1$

Explanation of Solution

Given:

Focus $\left(4,0\right)$

Directrix $x=2$

Calculation:

We are given the focus of hyperbola as $\left(4,0\right)$ and the directrix as $x=2$

The other parameters of the hyperbola can be calculated as:

  $\begin{array}{l}c=ae\\ =4\end{array}$

Also, eccentricity is

  $\begin{array}{l}\frac{a}{e}=2\\ \Rightarrow \frac{ae}{e^2}=2\\ \Rightarrow e^2=2\\ \Rightarrow e=\sqrt{2}\end{array}$

Now,

  $\begin{array}{l}PF=\sqrt{2}PD\\ \Rightarrow \sqrt{{\left(x-4\right)}^2+{\left(y-0\right)}^2}=\sqrt{2}\left|x-2\right|\\ \Rightarrow {\left(x-4\right)}^2+{\left(y-0\right)}^2=2{\left(x-2\right)}^2\\ \Rightarrow -x^2+y^2=-8\\ \Rightarrow \frac{x^2}{8}-\frac{y^2}{8}=1\end{array}$

Conclusion:

Therefore,the hyperbola’s standard-form equation is $\frac{x^2}{8}-\frac{y^2}{8}=1$

Eccentricity is $e=\sqrt{2}$