Answer
(a) Using the result ${r_{per}} = a\left( {1 - e} \right)$ from Exercise 13, we show that $C{F_1} = ae$
(b) Using $r = \frac{p}{{1 + e\cos \theta }}$ we show that
$a = \frac{p}{{1 - {e^2}}}$
(c) We obtain ${F_1}B = {F_2}B = a$. Thus, ${F_1}B + {F_2}B = 2a$.
(d) Using Pythagorean Theorem and the result from part (b) we prove that
$b = \sqrt {pa} $
Work Step by Step
(a) From Figure 10, we see that ${F_1}A = {r_{per}}$.
From Exercise 13, we obtain ${r_{per}} = a\left( {1 - e} \right)$. Thus,
$C{F_1} = CA - {F_1}A$
$C{F_1} = a - a\left( {1 - e} \right) = ae$
Hence, $C{F_1} = ae$.
(b) We have
$r = \frac{p}{{1 + e\cos \theta }}$, ${\ \ \ }$ where $p = {J^2}/k$.
So, at $\theta = 0$ we get ${r_{per}} = \frac{p}{{1 + e}}$.
From Exercise 13, we obtain ${r_{per}} = a\left( {1 - e} \right)$. Thus,
$a\left( {1 - e} \right) = \frac{p}{{1 + e}}$
$p = a\left( {1 - {e^2}} \right)$
Hence, $a = \frac{p}{{1 - {e^2}}}$.
(c) From part (a) we obtain $C{F_1} = ae$.
From Figure 10, we see that ${F_1}A = a - C{F_1}$ and ${F_2}A = {F_2}C + a$.
Since ${F_2}C = C{F_1}$, so ${F_2}A = C{F_1} + a$. Thus,
${F_1}A + {F_2}A = a - C{F_1} + C{F_1} + a$
Hence, ${F_1}A + {F_2}A = 2a$.
Since the $\vartriangle BC{F_1}$ is a right triangle, so
${F_1}B = \sqrt {{{\left( {CB} \right)}^2} + {{\left( {C{F_1}} \right)}^2}} $
${F_1}B = \sqrt {{b^2} + {a^2}{e^2}} = a\sqrt {\frac{{{b^2}}}{{{a^2}}} + {e^2}} $
Since $\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$, so ${F_1}B = a$.
By symmetry we obtain ${F_1}B = {F_2}B = a$. Thus, we conclude that ${F_1}B + {F_2}B = 2a$.
(d) By the Pythagorean Theorem:
$C{B^2} = {\left( {{F_1}B} \right)^2} - {\left( {C{F_1}} \right)^2}$
Since $CB = b$, ${F_1}B = a$, and $C{F_1} = ae$, so
${b^2} = {a^2} - {a^2}{e^2} = {a^2}\left( {1 - {e^2}} \right)$
$\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$
From part (b), we obtain $a = \frac{p}{{1 - {e^2}}}$. So,
$p = a\left( {1 - {e^2}} \right)$
$p = a\frac{{{b^2}}}{{{a^2}}} = \frac{{{b^2}}}{a}$
Hence, $b = \sqrt {pa} $.
