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Deriving Cartesian equation of a rectangular hyperbola

Another conics question (sorry!!)

I understand how to derive the Cartesian equation of a hyperbola (East-West) from the definition distance1 - distance2= constant=2a

The cartesian equation is: x^2/a^2 - y^2/b^2 = 1

Now, by definition a rectangualr hyperbola has eccentricty sqrt2.
Thus 2=c^2/a^2. Using c^2=b^2+a^2 identity for hyperbolas, this becomes
2= (b^2+a^2)/a^2, which gives that a=b.

This then means that the fundamental box for a standard hyperbola, becomes the fundamental square, with side 2a. Hence the asyptotes now become y=+or-x meaning that they are perpendicular to one another. This explains why a rectangular hyperbola has, by definition, perp. asymptotes.

I understand how if we make the x and y axes the asymptotes by rotation of pi/4, why get the familiar rect. hyperbola sketch. Am I right in saying that a rect. hyperbola can exist at any rotation with in the plane?

Now my question is, as rect. hyperbola are a type of hyperbola, then a rect. hyperbola Cartesian equation can surely be derived from that for a standard hyperbola:

x^2/a^2 - y^2/b^2 = 1

I am told that for rectangular hyperbolas in this orientation, that xy=c^2

so my question (eventually!!) is how does one get from standard cartesian equation to xy=c^2?

I would prefer as thorough an explanation as possible.

Thank you!

watching this thread

I'd be keen to find an answer to this question too :smile:

Reply 3

Erebusaur

Right. Okay, so you start with your East-West cartesian equation:

[br]\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1[br]

As you've correctly deduced, a = b and we thus get this equation:

[br]x^{2} - y^{2} = a^{2}[br]

In order to rotate from the rectangular hyperbola to the East-West hyperbola, we need to rotate by -pi/4 to get back to the (x,y) variables used above:

[br]\begin{pmatrix}[br]x\\ [br]y[br]\end{pmatrix}=\begin{pmatrix}[br]\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ [br]-\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}[br]\end{pmatrix}\begin{pmatrix}[br]x'\\ [br]y'[br]\end{pmatrix}=\begin{pmatrix}[br]\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'\\ [br]-\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'[br]\end{pmatrix}[br]

In other words:

[br]x = \dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'[br]y = -\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'[br]

But these must satisfy our equation above, so we get:

Unparseable latex formula:

[br]x^{2} = \left (\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'\right )\left (\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'\right )= \dfrac{x'^{2}}{2} + x'y' + \dfrac{y'^{2}}{2}}[br]

and:

Unparseable latex formula:

[br]y^{2} = \left (-\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'\right )\left (-\dfrac{\sqrt{2}}{2}x' + \dfrac{\sqrt{2}}{2}y'\right )= \dfrac{x'^{2}}{2} - x'y' + \dfrac{y'^{2}}{2}}[br]

So:

[br]x^{2} - y^{2} = \left (\dfrac{x'^{2}}{2} + x'y' + \dfrac{y'^{2}}{2}\right ) - \left (\dfrac{x'^{2}}{2} - x'y' + \dfrac{y'^{2}}{2}\right ) = 2x'y'[br]\therefore 2x'y' = a^{2}[br]\therefore x'y' = \dfrac{a^{2}}{2} = c^{2}[br]

I'm sorry but I'm too tired to fill this out with words. I'll try to do that tomorrow morning. I hope this helps anyway.

(edited 9 years ago)

Reply 4

TeeEm

Original post by Choochoo_baloo

Another conics question (sorry!!)

...

Original post by 2011wc2013ct

watching this thread

Original post by Ensign123

I'd be keen to find an answer to this question too :smile:

Original post by Erebusaur

...
I

Erebusaur explained it already (very good LaTex skills)

I will also attach my personal conic section reference which has the same proof at the end which you may find useful

CONIC SECTION REFERENCE.pdf

Reply 5

harmingjuli

What module is this?