How to create a Hyperbola curve using Equation Curve (Autodesk Inventor 2013)

[SORJM]

I want to draw hyperbola curve whose equation is  x^2/a^2  - y^2/b^2=1. May I use equation curve tool to make the above hyperbola.  If it is possible please help me to do so. 

I have also made a hyperbola curve of aforesaid equation through conventional method on mathematical ground in the past. I am posting the file of the same to make the question clearer. 

Regards

SORJM

[karthur1 · Accepted solution]

First thing, write your equation in terms of Y=SQRT(b^2((x/a)^2-1)). For the equation curve, set x=t

Kirk A.

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[SORJM]

Thanks for the quick and valuable response. It shows your excellent command over the software. 

Regards

SORJM

[karthur1]

Here is another example of an hyperbola with these parameters.

Center  (420 , 0) mm
 a = 300mm
 b = 80mm

So the equation is:  

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

[Anonymous]

Is it possible to draw a hyperbola curve from a sheet of metal by the given values?

a=1.9

b=1.46

c=2.4

 and the following equation is x²/a² – y²/b² = 1 

Regards,

Amjad

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@karthur1 wrote:

Here is another example of an hyperbola with these parameters.

Center  (420 , 0) mm
 a = 300mm
 b = 80mm

So the equation is:  

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

---

[karthur1]

Your equation is in the form X^2-Y^2=1    (X^2/a^2) = (X/a)^2.... which is like the example in message 5 I posted.

If you still cant get it, post your ipt and I will see what I can do.

[Anonymous]

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@karthur1 wrote:

Your equation is in the form X^2-Y^2=1    (X^2/a^2) = (X/a)^2.... which is like the example in message 5 I posted.

If you still cant get it, post your ipt and I will see what I can do.

---

[karthur1]

Here is the equation for what you asked for. a=1.9, b=1.46

Not sure where to put the "C" value you gave.

Here is the results after I mirrored the initial curve a couple times.  See attached file (Inv Version 2018)

[karthur1]

Since the equation that I worked out if for a vertical hyperbola, I had to rotate the axis for your problem.  To do that, I had to change some things around

Where a=1.90 and b=1.46, I then have this for the equation curve.

See attached file of the part.