Chapter 8.5, Problem 58E

Solution

a)

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

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

Given information:

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

  $c^2=a^2-b^2$

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

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

Coordinates: $P\left(x,y\right)$ , $D\left(\frac{a^2}{c},y\right)$ , $F\left(c,0\right)$

Formula used:

Distance formula: $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

To find the distance $PD$ :

Substitute $\frac{a^2}{c}$ for $x_2$ , $y$ for $y_1$ and $y_2$ .

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

Substitute $c$ for $x_2$ , $x$ for $x_1$ , $y$ for $y_1$ and $0$ for $y_2$ .

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

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 $y^2$ .

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

Multiply both sides of the equation by

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

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

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

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

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

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

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

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

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

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

b)

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

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

Given information:

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

  $c^2=a^2-b^2$

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

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

Coordinates: $P\left(x,y\right)$ , $D\left(\frac{a^2}{c},y\right)$ , $F\left(c,0\right)$

Calculation:

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

Hence, $e=\frac{c}{a}$

c)

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

It is proved that the distance from $F$ to $L$

  $FL=\frac{a}{e}-ae$ .

Given information:

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

  $c^2=a^2-b^2$

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

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

Coordinates: $P\left(x,y\right)$ , $D\left(\frac{a^2}{c},y\right)$ , $F\left(c,0\right)$

Formula used:

  $e=\frac{c}{a}$

Calculation:

  $FL=\frac{a^2}{c}-c$

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 $c$ in $FL$ .

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

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