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I have this problem:

Find the canonical equation of an hyperbola if the distance between the directrices is $\frac{8}{3}$ and the eccentricity $e=\frac{3}{2}$.

How would you solve it?

This is my try:

The canonical equation of a hyperbola takes the form $x^2/a^2 - y^2/b^2 = 1$, and the foci are at a distance $c > a$ from the origin, and our directrices are located at $a^2/c$, where eccentricity is $\sqrt{a^2+b^2}/a$. If $e = 3/2$, then

\begin{align} \frac{\sqrt{a^2+b^2}}{a} = \frac32 &\implies \frac32 a = \sqrt{a^2+b^2} \\[4pt] &\implies \frac94 a^2 = a^2 + b^2 \\[4pt] &\implies a^2\left(\frac94 - 1\right) = b^2 && (c^2 = a^2 + b^2) \end{align}

so \begin{align} a^2 + a^2\left(\frac94-1\right) = c^2 &\implies c^2 = a^2\left(1 + \frac94 - 1\right) \\[4pt] &\implies c^2 = \frac94 a^2 \\[4pt] &\implies c = \frac32 a \end{align} so our directrix is located at

$$\frac{a^2}{c} = \frac{a^2}{a\cdot 3/2} = \frac{a}{3/2} = \frac{2}{3}\cdot a,$$ but the distance between directrices is $8/3$, so it's double the distance from the origin, so essentially, $$ \frac{8}{3} = 2x = 2\cdot \frac{2}{3} \cdot a \implies \frac{8}{3} = \frac{4}{3} \cdot a \implies a = 2. $$ And since

$$ c^2 - a^2 = b^2 = \frac{9}{4} \cdot 4^2 - 4^2 = 16 \left( \frac{9}{4} - 1\right) = 4 \cdot 9 - 16 = 36-16 =20. $$

In total, we have $$a^2 = 4, \qquad\text{and}\qquad b^2 = 20,$$ which results in the canonical form $$\frac{x^2}{4} -\frac{y^2}{20} = 1.$$

Also, how would you find the eccentricity of an ellipse if the sides of the square inscribed in it pass through the foci of the ellipse?

My try:

Let’s imagine a generic ellipse, and the square inside has its sides located at $-x$ and $x$, making the side of the square $2x$, and we know that the foci occur at $$ x = \frac{a^2}{c} \quad\text{so}\quad 2x = \frac{2a^2}{c}, \qquad\text{(since $c^2 = a^2 + b^2$)},$$ and eccentricity is $$ \frac{\sqrt{a^2 + b^2}}{a} = \frac{c}{a}, $$ so if $ e = c/a $ and $$ 2x = \frac{2a^2}{c} = 2a \cdot \frac{a}{c} = 2a\cdot \frac{1}{e} = \frac{2a}{2} \implies x = \frac{a}{2} \implies e = \frac{a}{x} $$ and $x$ would be half the side of the square in this case.

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    $\begingroup$ What have you tried? $\endgroup$ Commented Jul 10, 2020 at 5:35
  • $\begingroup$ It would help to add what you've tried so far / how far you've gotten. Not often will people receive an answer with no work done prior. $\endgroup$
    – Graviton
    Commented Jul 10, 2020 at 5:35
  • $\begingroup$ Welcome to Math.SE! ... You should ask the ellipse question separately. ... In any case, questions should include something of what you know about the problem. (What have you tried? Where did you get stuck? etc) This helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know or using techniques beyond your skill level. (It also helps convince people that you aren't simply trying to get them to do your homework for you. Isolated problem statements tend to give the wrong impression in this regard.) $\endgroup$
    – Blue
    Commented Jul 10, 2020 at 5:36
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    $\begingroup$ Pure problem statement are considered off-topic here and may be flagged for closure. Can you add more information of the context like where did you encountered this question or what more do you know about the mathematical terms you used. It will be helpful for other users also to directly address the problem. $\endgroup$
    – SarGe
    Commented Jul 10, 2020 at 6:46
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    $\begingroup$ Thanks for adding your attempt, and please wait for your question to be reopened. $\endgroup$
    – user21820
    Commented Jul 18, 2020 at 15:21

1 Answer 1

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In the hyperbola exercise, you double-squared the $a$ in your $c^2-a^2 = \cdots = 20$ calculation, effectively multiplying everything by an extra factor of $4$. You should get $9-4=5$, instead.


Here's a clearer path to the solution:

If the hyperbola's transverse semi-axis is $a$, its center-to-focus distance is $c$, and its eccentricity is $e$, then the center-to-directrix distance (call it $d$) is indeed given by $d=a^2/c$, so that (since $e=c/a$) we can write $a = de$.

Since the distance between the directrices is $8/3$, we have $d=4/3$; given the eccentricity $3/2$, we have $$a = de=\frac43\cdot\frac32=2 \tag{1}$$ Then, $$c=ae = 2\cdot\frac32=3 \tag{2}$$ and then $$b^2=c^2-a^2=9-4=5\tag{3}$$ so that, for an origin-centered hyperbola with a horizontal transverse axis, the equation is $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad\to\quad \frac{x^2}{4}-\frac{y^2}{5}=1 \tag{$\star$}$$


(I think the ellipse question should be posted separately, so I won't address it here.)

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