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The four Latusrectum of two hyperbolas forms four distinct sides of a rectangle. If $e,E$ are their eccentricities, then which of the following are true?

  1. $eE\geq 2$

  2. $\left(e-\dfrac{1}{e}\right)\left(E-\dfrac{1}{E}\right)=1$

  3. $eE\geq 3$

  4. both hyperbolas can never have same eccentricities

This is a question in a test of my student. We knew the answers 1), 2) are correct. But couldn't figure out why?

I have tried to make figures, but converting into algebraic expressions is difficult.

It would be a great pleasure if someone helps!

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3 Answers 3

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Let the first hyperbola be

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, with eccentricity $e$.

And the second hyperbola be

$\dfrac{y^2}{B^2} - \dfrac{x^2}{A^2} = 1$ with eccentricity $E$.

Then the right focus of the first hyperbola is at $x = a e$

and the top focus of the second hyperbola is at $y = B E$

The y-coordinate of the upper end of the latus rectum of the first hyperbola is found by substituting $x = a e$ into the the equation of the hyperbola

$e^2 - \dfrac{y^2}{b^2} = 1$

so,

$y = b \sqrt{e^2 - 1}$

and this is equal to $BE$

i.e. $ B E = b \sqrt{e^2 - 1} $

and similarly

$ a e = A \sqrt{E^2 - 1}$

hence

$ B = \dfrac{b}{E} \sqrt{e^2 - 1}$

$ A = \dfrac{a e}{ \sqrt{E^2 - 1} }$

and we know that $E^2 = 1 + \dfrac{A^2}{B^2}$

so

$E^2 B^2 = A^2 + B^2 = \dfrac{b^2 (e^2 - 1)}{E^2} + \dfrac{a^2 e^2}{ (E^2 - 1)} $

But $E^2 B^2 = b^2 (e^2 - 1)$, therefore,

$b^2 (e^2 - 1) = \dfrac{b^2 (e^2 - 1)}{ E^2} + \dfrac{a^2 e^2}{(E^2 - 1)}$

Collecting terms,

$b^2 (e^2 - 1) ( 1 - \dfrac{1}{E^2}) = \dfrac{a^2 e^2}{(E^2 - 1)}$

now $e^2 = \dfrac{ (a^2 + b^2) }{ a^2 }$

hence

$\dfrac{b^2 (e^2 - 1) (E^2 - 1)}{ E^2 } = \dfrac{(a^2 + b^2)}{ (E^2 - 1)}$

Cross multiplying, and dividing through by $b^2$

$ \dfrac{(E^2 - 1)^2}{E^2} = \dfrac{(a^2 + b^2)}{( b^2 (e^2 - 1) ) }= \dfrac{( (a/b)^2 + 1 ) }{ (e^2 - 1) } $

but $\dfrac{b^2}{a^2} = e^2 - 1$, so $\dfrac{a^2}{b^2} = \dfrac{1}{(e^2 - 1)}$ , and $\dfrac{a^2}{b^2} + 1 = \dfrac{e^2 }{(e^2 - 1)}$

Hence,

$\dfrac{(E^2 - 1)^2}{E^2} = \dfrac{e^2}{(e^2 - 1)^2} $

Taking the square root,

$\dfrac{(E^2 - 1)}{E} = \dfrac{e}{(e^2 - 1)}$

thus

$(E - \dfrac{1}{E})(e - \dfrac{1}{e}) = 1 $

This proves that #2 is true.

Let's check if #4 is true or not. If $E = e$ then

$(e - \dfrac{1}{e})^2 = 1$

so

$e^2 + \dfrac{1}{e^2} - 2 = 1$

i.e. $e^2 + \dfrac{1}{e^2 }= 3$

or $(e^2)^2 - 3 (e^2) + 1 = 0$

and this gives

$e^2 = \dfrac{1}{2} ( 3 \pm \sqrt{5} )$

since $e \gt 1$ , then $e^2 = \dfrac{1}{2} (3 + \sqrt{5}) \approx 2.618 $

Therefore #4 is False.

Finally, we want to check the range for $eE$. We now know that

$(E^2 - 1)(e^2 - 1) = E e$

Hence,

$(E e)^2 - e^2 - E^2 + 1 = E e define $u = E e$, and $v = \dfrac{E}{ e} $

then $ u v = E^2 $ and $ \dfrac{u}{v} = e^2 $

Hence,

$u^2 = u v + \dfrac{u}{ v} + u - 1 = u ( 1 + v + \dfrac{1}{v}) - 1 \gt 3 u - 1 $

Therefore, we now have,

$ u^2 - 3 u + 1 \gt 0 $ where $u > 1$

The roots of $u^2 - 3u + 1 = 0$ are $u \approx 0.3819$ and $ u \approx 2.618 $. Since u satisfies the above inequality , and $u \gt 1$ then $u \gt 2.618 $

hence $u \gt 2$, however $u$ is not necessarily greater than $3$. So #1 is true,but #3 is False.

An example of the given situation is depicted below with $e = 1.5$ and $a = 1 $

enter image description here

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Assuming the equations of the 2 hyperbolas as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, eccentricity= e and $-\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$,(E). the rectangle will be made by lines $x=\pm ae$ and $y=\pm BE$ (equations of the 4 latus rectums). the coordinates of the rectangle will be (ae,BE),(ae,-BE),(-ae,-BE),(-ae,BE). The centre being the origin. The diagonals of the rectangle are also the asymptotes of the hyperbolas.

The slope of the rectangle is $m=\frac{BE}{ae}$ this m should be equal to tan45. So m=1 and ae=BE. The equation of diagonal: (y+BE) = 1(x+ ae) $\Rightarrow$ x-y=0. $\Rightarrow$ y=+x, which is an asymptote of rectangular hyperbola. So e must be $\sqrt2$. E may or may not be equal to $\sqrt2$ as E=ae/B and it depends on a/B and it isn't necessary for a and B to be equal.

Thus the 1st option is correct, with the limiting case in which both have eccentricities equal to $\sqrt2$, and value of E varying. While the 2nd option can't be satisfied by keeping the same values of e and E. When value of e = $\sqrt2$ is put, we get 2 values of E. One of which is acceptable (greater than 1). The 2nd option then shouldn't be wrong.

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  • $\begingroup$ Welcome to Math.SE! ... Unfortunately, your answer has a few errors: "The diagonals of the rectangle are also the asymptotes of the hyperbolas." This is not so. The corners of the rectangle contain the endpoints of the latera recta; these lie on the hyperbolas, not on their asymptotes. ... "The slope of the rectangle is $m=\frac{BE}{ae}$ this m should be equal to tan45" I'm guessing that by "slope of the rectangle" you mean "slope of a diagonal of the rectangle". Even if so, that $m$ is only $\tan45^\circ=1$ if the rectangle is a square, which isn't necessarily the case. $\endgroup$
    – Blue
    Apr 3 at 8:38
  • $\begingroup$ Thanks to Xyz. Really it's a Great effort. Thank you so much. $\endgroup$
    – YBR
    Apr 4 at 8:05
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Recall that in a hyperbola with transverse radius $a$, conjugate radius $b$, and eccentricity $e := \dfrac{\sqrt{a^2+b^2}}{a}$, the center-to-focus distance is $c:=\sqrt{a^2+b^2}=ae$, and the semi-latus rectum has length $$\frac{b^2}{a}=\frac{a^2e^2-a^2}{a}=a(e^2-1)$$

Consider, then, hyperbola with transverse radii $p$ and $q$, and respective eccentricities $e$ and $f$, whose latera recta form a rectangle, as shown:

enter image description here

The half-height of the rectangle is both the semi-latus rectum of the $p$-hyperbola and the center-to-focus length of the $q$-hyperbola. Comparably for the half-width. Therefore, we have $$ \left.\begin{array}{l} p\left(\,e^2-1\right)=qf \\ q\left(f^2-1\right)=pe \end{array}\right\} \quad\to\quad\frac{p}{q}=\frac{f}{e^2-1}=\frac{f^2-1}{e} \tag{A} $$

We'll show that the question's options (1) and (2) are true, while $(3)$ and $(4)$ are false.

  • Equation $(A)$ implies $$\left(e^2-1\right)\left(f^2-1\right) = ef \quad\to\quad \frac{e^2-1}{e}\cdot\frac{f^2-1}{f} = 1 \quad\to\quad \left(e-\frac1e\right)\left(f-\frac1f\right) = 1 \tag{B}$$ which proves the property in option $(2)$.

  • For option (1), we can re-write the first equation in $(B)$ as $$e f\;(e f-2) \;=\; (e f-1)\;+\;(e-f)^2 \tag{C}$$ Since both $e$ and $f$ exceed $1$ as eccentricities of hyperbolas, and since $(e-f)^2$ is necessarily non-negative, the right-hand side of $(C)$ must be strictly positive; hence, the left-hand side must be as well, and we conclude that $ef>2$. This makes the question's option $(1)$ true, but slightly over-generous in allowing the possibility of $ef=2$ (which never actually occurs).

  • As for options $(3)$ and $(4)$: Note that $e=f$ implies, from $(A)$, that $\dfrac{p}{q}=\dfrac{q}{p}$, so that $p=q$, making the hyperbolas congruent and the rectangle a square. Moreover, since $p/q=1$, we can solve (ignoring the negative root) $$e^2-1=e \quad \to \quad e = \frac12\left(1+\sqrt{5}\right) = 1.618\ldots \tag{D}$$ which makes the eccentricity the Golden Ratio, $\phi$. That this exceeds $1$ makes it a valid hyperbola eccentricity, so that option (4)'s assertion that the eccentricities cannot match *false; moreover, since $ef=\phi^2=1+\phi=2.618\ldots$ is less than $3$, option $(3)$ is also false.

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