Chapter 8.5, Problem 59E

Solution

a)

To prove: that $L$ is a directrix for the hyperbola.

It is proved that $L$ is a directrix for the hyperbola.

Given information:

  

Equation of hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci: $\left(\pm c,0\right)$

Half the length of the transverse axis is $a$ .

  $x=\frac{a^2}{c}$

Formula used:

  $c^2=a^2+b^2$

Algebraic identity: ${\left(a+b\right)}^2=a^2+b^2+2ab$

Calculation:

  $\begin{array}{c}PF=\sqrt{{\left(x-c\right)}^2+{\left(y-0\right)}^2}\\ =\sqrt{{\left(x-c\right)}^2+y^2}\end{array}$

  $\begin{array}{c}PD=x-\frac{a^2}{c}\\ =\frac{cx-a^2}{c}\end{array}$

Substitute $PF$ and $PD$ in ratio.

  $\frac{PF}{PD}=\frac{c\sqrt{x^2+c^2-2cx+y^2}}{a^2-cx}$

The equation of the ellipse is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

Solve for $y^2$ .

  $\begin{array}{c}\frac{y^2}{b^2}=\frac{x^2}{a^2}-1\\ =\frac{x^2-a^2}{a^2}\end{array}$

Multiply both sides of the equation by

  $y^2=\frac{b^2\left(x^2-a^2\right)}{a^2}$

Substitute $\frac{b^2\left(x^2-a^2\right)}{a^2}$ for $y^2$ in $\frac{c\sqrt{x^2+c^2-2cx+y^2}}{cx-a^2}$

  $\begin{array}{l}\frac{PF}{PD}=\frac{c\sqrt{x^2+c^2-2cx+\frac{b^2\left(x^2-a^2\right)}{a^2}}}{cx-a^2}\\ =\frac{c\sqrt{\frac{a^2{x}^2+a^2{c}^2-2a^{2}cx+b^2\left(x^2-a^2\right)}{a^2}}}{cx-a^2}\\ =\frac{c\sqrt{a^2{x}^2+a^2{c}^2-2a^{2}cx+b^2\left(x^2-a^2\right)}}{cx-a^2}\end{array}$

Substitute $c^2-a^2$ for $b^2$ .

  $\begin{array}{l}=\frac{c\sqrt{a^2{x}^2+a^2{c}^2-2a^{2}cx+\left(c^2-a^2\right)\left(x^2-a^2\right)}}{a\left(cx-a^2\right)}\\ =\frac{c\sqrt{a^2{x}^2+a^2{c}^2-2a^{2}cx+c^2{x}^2-c^2{a}^2-a^2{x}^2+a^4}}{a\left(cx-a^2\right)}\end{array}$

Cancel out $+a^2{c}^2$ and $-a^2{c}^2$ . Cancel out $+a^2{x}^2$ and $-a^2{c}^2$ .

  $\begin{array}{l}=\frac{c\sqrt{-2a^{2}cx+a^4+c^2{x}^2}}{a\left(cx-a^2\right)}\\ =\frac{c\sqrt{{\left(cx-a^2\right)}^2}}{a\left(cx-a^2\right)}\end{array}$

As $\sqrt{a^2}=a$ , $\sqrt{{\left(cx-a^2\right)}^2}=\left(cx-a^2\right)$ .

  $\begin{array}{l}=\frac{c\left(cx-a^2\right)}{a\left(cx-a^2\right)}\\ =\frac{c}{a}\end{array}$

Therefore, $L$ is a directrix for the hyperbola.

b)

To prove: that the eccentricity is $e=\frac{c}{a}$ .

It is proved that $e=\frac{c}{a}$ .

Given information:

  

Equation of hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci: $\left(\pm c,0\right)$

Half the length of the transverse axis is $a$ .

  $x=\frac{a^2}{c}$

Calculation:

The eccentricity is the ratio of the distance from focus to the distance from directrix. This is proved in part (a).

  $\begin{array}{l}e=\frac{PF}{PD}\\ e=\frac{c}{a}\end{array}$

Hence, it is proved that $e=\frac{c}{a}$ .

c)

To prove: that the distance from $F$ to $L$ is $a/e-ea$

It is proved that $FL=ae-\frac{a}{e}$ .

Given information:

  

  $L$ is the vertical line $x=\frac{a^2}{c}$ .

Formula used:

Formula to find the distance between two points: $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Substitute $\frac{a^2}{c}$ for $x_1$ , $c$ for $x_2$ , and $0$ for both $y_1$ and $y_2$ in the distance formula.

  $\begin{array}{c}FL=\sqrt{{\left(c-\frac{a^2}{c}\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(c-\frac{a^2}{c}\right)}^2}\\ =c-\frac{a^2}{c}\end{array}$

Eccentricity is proved to be $e=\frac{c}{a}$ .

Solve for $c$ .

  $\begin{array}{l}e=\frac{c}{a}\\ ae=c\end{array}$ ,

Substitute $ae$ for $c$ .

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

Therefore, $FL=ae-\frac{a}{e}$