# Relationships between the eccentricities of two hyperbolas whose latusrectums form a rectangle

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!

## 3 Answers

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 $$a = 1$$ 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. • 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.
– Blue
Apr 3 at 8:38
• Thanks to Xyz. Really it's a Great effort. Thank you so much.
– YBR
Apr 4 at 8:05

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:

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.