Different types of hyperbolas and their properties

[Gourab_chill]

Homework Statement

The question along the answer is given in the attachments. What I seek is the explanation.

Relevant Equations

---not sure---

I know the hyperbola of the form x^2/a^2-y^2/b^2=1 and xy=c; but coming across this question I'm put in a dilemma of how to proceed with calculating anything of it - say eccentricity or latus rectum or transverse axis as said. How to generalize a hyperbola (but i don't want a complex derivation of hyperbolas because that's out of my understanding, i want to get across these questions; teach me the fundamentals) and find it's different values or parameters?

 

[BvU]

Hi,

but i don't want a complex derivation of hyperbolas

Fair enough. So no references like this ?
In that case you can fall back to a very useful action: make a sketch like I did:

Blue is your hyperbola, red is the asymptotes and on the brown line is the transverse axis.

With such a nice sketch it is clear what you need to do to calculate the required value !

--

Other approach: calculus.

 

[Gourab_chill]

@BvU looks neat; but here is the problem how do I know what is the equation of the transverse axis? And how do I calculate the quantities like eccentricity, latus rectum length? What should be my approach?

 

[BvU]

equation of the transverse axis

You know the equations of the asymptotes

And I admit the others are beyond my recollection (if I even ever encountered them). Will you look them up or should we do that for you ?

However, might be there's no escaping

complex derivation

then

 

[Physics lover]

Ok so I will give you a hint.You shall proceed further.Since the given equation contains xy term,therefore it is clear that the hyperbola has been rotated.
So to get back to the original form of hyperbola,substitute
X=xcosA+ysinA
Y=xcosA-ysinA
or any other substitution you prefer.
Now,you must be able to proceed further.

 

[haruspex]

how do I know what is the equation of the transverse axis?

You are not asked for the equation, only the length.
You can see from @BvU's beautiful sketch that it's the closest approach to the origin. Just a little calculus...

 

[BvU]

@Gourab_chill : did you find the asymptote equations ? Then you can also find the line for the transverse axis and then finding the marked point doesn't even require any calculus !

beautiful sketch

too much credit -- just Excel

 

[Physics lover]

Ok so I will give you a hint.You shall proceed further.Since the given equation contains xy term,therefore it is clear that the hyperbola has been rotated.
So to get back to the original form of hyperbola,substitute
X=xcosA+ysinA
Y=xcosA-ysinA
or any other substitution you prefer.
Now,you must be able to proceed further.

Is this not the correct way @haruspex @BvU

 

[SammyS]

Ok so I will give you a hint.You shall proceed further.Since the given equation contains xy term,therefore it is clear that the hyperbola has been rotated.
So to get back to the original form of hyperbola,substitute
X=xcosA+ysinA
Y=xcosA-ysinA
or any other substitution you prefer.
Now,you must be able to proceed further.

Is this not the correct way @haruspex @BvU

Not quite.

Change the equation for Y. The pair of equations becomes:

$X=x\cos A+y\sin A$
$Y=y\cos A-x\sin A$

Perhaps this is what you intended.

 

[BvU]

Then you can also find the line for the transverse axis

The brown line is a bit too suggestive, going through $(1,-0.5)$.

View attachment 261931

The green dash-dot lines represent the correct axes:

So you'll need an intersection of the green line and the hyperbola.

Not by coincidence the slope of that green line is $\ -\tan(A)\$ in @SammyS post #9

 

[haruspex]

Is this not the correct way @haruspex @BvU

There is no one correct way. Even what is the easiest way may depend on what the solver is most comfortable with.
To me, what I outlined in post #6 seemed easier than figuring out the rotation.

 

[Physics lover]

Not quite.

Change the equation for Y. The pair of equations becomes:

$X=x\cos A+y\sin A$
$Y=y\cos A-x\sin A$

Perhaps this is what you intended.

yes.That was a typo.Sorry for that.

 

[Charles Link]

yes.That was a typo.Sorry for that.

@Physics lover Your post is a good one. Especially for the case of $xy=1$, the best way of looking at it IMO is the rotation of axes by 45 degrees which does put it in the standard form: $\frac{x'^2}{a^2}-\frac{y'^2}{b^2}=1$, with $a=b=\sqrt{2}$. $\\$ Without this result, someone might think that there are different types of hyperbolas. I believe they can all be put in the standard form with the necessary rotation and/or translation.

 

[haruspex]

@Physics lover Your post is a good one. Especially for the case of $xy=1$, the best way of looking at it IMO is the rotation of axes by 45 degrees which does put it in the standard form: $\frac{x'^2}{a^2}-\frac{y'^2}{b^2}=1$, with $a=b=\sqrt{2}$. $\\$ Without this result, someone might think that there are different types of hyperbolas. I believe they can all be put in the standard form with the necessary rotation and/or translation.

I still favour
$x^2+y^2=2x^2-2+\frac 1{x^2}=2u-2+\frac 1u$, where u=x².
Then it is just a matter of finding the minimum of that and taking the square root.

 

[Charles Link]

I still favour
$x^2+y^2=2x^2-2+\frac 1{x^2}=2u-2+\frac 1u$, where u=x².
Then it is just a matter of finding the minimum of that and taking the square root.

@haruspex Yes, your way is easier, but I successfully worked it by considering the rotation. It took a little work to get the answer given by the OP, but I got it.

 

[SammyS]

@Physics lover Your post is a good one. Especially for the case of $xy=1$, the best way of looking at it IMO is the rotation of axes by 45 degrees which does put it in the standard form: $\frac{x'^2}{a^2}-\frac{y'^2}{b^2}=1$, with $a=b=\sqrt{2}$. $\\$ Without this result, someone might think that there are different types of hyperbolas. I believe they can all be put in the standard form with the necessary rotation and/or translation.

The rotation is 22.5° .

 

[haruspex]

The rotation is 22.5° .

I think @Charles Link was just applying it to the example of xy=1, not the hyperbola in the question.

 

[ehild]

I think @Charles Link was just applying it to the example of xy=1, not the hyperbola in the question.

@Charles Link showed a way of solution for the problem in the OP. And the hyperbola is really rotated by 22.5 degrees clockwise with respect to a hyperbola of standard from x²/a²-y²/b²=1

 

[BvU]

@Charles Link showed a way of solution for the problem in the OP. And the hyperbola is really rotated by 22.5 degrees clockwise with respect to a hyperbola of standard from x²/a²-y²/b²=1

Yes, the green line in the picture in #10.

I was surprised that the tangent of $\pi/8$ has a value $\ \sqrt2 - 1 \$ (wasn't in my dusty brain archive )

In our enthousiasm, have we lost @Gourab_chill ? What's the status with you ?

https://www.mathsisfun.com/geometry/hyperbola.html

 

[haruspex]

@Charles Link showed a way of solution for the problem in the OP. And the hyperbola is really rotated by 22.5 degrees clockwise with respect to a hyperbola of standard from x²/a²-y²/b²=1

That's not how I read Charles' post.
He endorsed the rotation method, but then illustrated it for a different hyperbola, xy=1, for which the rotation is 45°. @SammyS seems to have misread this as saying that the rotation in the thread problem is 45°. But we might have to wait for Charles to confirm or deny.

 

[ehild]

I was surprised that the tangent of $\pi/8$ has a value $\ \sqrt2 - 1 \$ (wasn't in my dusty brain archive )

But your brain archive

certainly has sin, cos or tan of $\pi/4$ and you can derive $\tan(\pi/8)$ easily from the double angle formula
$$\tan(\pi/4)=\frac{2\tan(\pi/8)}{1-\tan^2(\pi/8)}=1$$
or the half-angle formulas
$$\sin(\pi/8)=\sqrt{\frac{1-\cos(\pi/4)}{2}}$$ and
$$\cos(\pi/8)=\sqrt{\frac{1+\cos(\pi/4)}{2}}$$

 

[Charles Link]

That's not how I read Charles' post.
He endorsed the rotation method, but then illustrated it for a different hyperbola, xy=1, for which the rotation is 45°. @SammyS seems to have misread this as saying that the rotation in the thread problem is 45°. But we might have to wait for Sammy to confirm or deny.

When I first did my post responding to @Physics lover and showed the case of $xy=1$, (and 45 degree rotation), I hadn't yet read the thread carefully enough to see the OP had an attachment that contained the more complicated $y=x-\frac{1}{x}$. $\\$ I later solved the OP's problem, by computing $\cos{\theta}$ so that the $x'y'$ term vanishes, and then finding where the line $y=(\tan{\theta})( x )$ intersects the curve $y=x-\frac{1}{x}$, and finally computing the distance of that point from the origin. @haruspex does have a simpler solution.

 

[neilparker62]

The brown line is a bit too suggestive, going through $(1,-0.5)$.

View attachment 261931

The green dash-dot lines represent the correct axes:

View attachment 261965

So you'll need an intersection of the green line and the hyperbola.

Not by coincidence the slope of that green line is $\ -\tan(A)\$ in @SammyS post #9

Based on above diagram, if equation of hyperbola is f(x), solve simultaneously:

y=f(x)
y=-x/f'(x)

ie: f(x) = -x/f'(x)

I think Haruspex went the same way (largely) (?)

 

[Charles Link]

You can solve $y=f(x)$ and $y=-tan(A) \, x$, but
@haruspex assumes the graph originates from the origin. He takes the distance $s$ as $s^2=x^2+y^2$ and cleverly substitutes for $y$ with $y=x-\frac{1}{x}$, so that he has $s^2$ as a function of $x$, (see post 14). He then minimizes $s^2$, obtaining the value for $x$ where the derivative is zero. Next, solve for $y$, and compute $s$.

 

[neilparker62]

You can solve $y=f(x)$ and $y=-tan(A) \, x$, but
@haruspex assumes the graph originates from the origin. He takes the distance $s$ as $s^2=x^2+y^2$ and cleverly substitutes for $y$ with $y=x-\frac{1}{x}$, so that he has $s^2$ as a function of $x$, (see post 14). He then minimizes $s^2$, obtaining the value for $x$ where the derivative is zero. Next, solve for $y$, and compute $s$.

Oh - ok. I am solving by assuming the minimized line passes through the origin and is perpendicular to the hyperbola.

 

[haruspex]

@haruspex assumes the graph originates from the origin.

I didn’t assume it, but neither did I include that part of the proof.
If you switch x to -x and y to -y in the equation for the curve you get the same equation back. Therefore it is symmetric about the origin.

 

[neilparker62]

Both methods need that proof.

 

[Charles Link]

@neilparker62 For your method, (with the curve being perpendicular to the line), you can solve $(f'(x))(-\tan(A))=-1$, which also gets a correct solution.

 

[Charles Link]

Just an add-on to the "proof" (see post 26) by @haruspex . What follows is essentially the same thing, but it might make it easier to follow: $\\$ Let $(x_{old}, y_{old} )$ be a point on the curve. It satisfies $y_{old}=x_{old}-\frac{1}{x_{old}}$. $\\$ Let $x_{new}=-x_{old}$. We have the corresponding $y_{new}=x_{new}-\frac{1}{x_{new}}=-x_{old}+\frac{1}{x_{old}}=-y_{old}$, so that $y_{new}=-y_{old}$. Thereby, in general, if $(x,y)$ is on the curve, $(-x, -y)$ is also on the curve. (It may be a separate branch of the curve, but it's still the "curve"). $\\$ Getting the original equation with the substitution $x=-x$ and $y=-y$ is essentially the same thing, but the reader might find this alternative method to be useful.

 

[Gourab_chill]

In our enthousiasm, have we lost @Gourab_chill ? What's the status with you ?

https://www.mathsisfun.com/geometry/hyperbola.html

thanks for the answers :)