Rally easy question (but I still )

[mmh37]

I am trying to draw the contour plot of the following function:

$$f(x,y) = \frac {a(x+y)} {x^2 + y^2 + a^2}$$

by setting h(x,y) = 1 and completing the square I received the following:

$$(x-(a/2))^2 - (y-a/2)^2 = a^2$$

surely, this is not a circle, and not a ellipse...but what is it then.

Also, is there a possibility to draw contour plots rather fast (e.g. with Mathcad)?

 

[quasar987]

Hyperboleas are caracterised by the implicit relation

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

 

[mmh37]

but I thought there would be a + in between the square brackets in order for it to be a circle, i.e.

$$(x-(a/2))^2 + (y-a/2)^2 = a^2$$

or does it simply not matter?

 

[quasar987]

but I thought there would be a + in between the square brackets in order for it to be a circle, i.e.

$$(x-(a/2))^2 + (y-a/2)^2 = a^2$$

or does it simply not matter?

You read my post before I could delete it. I had missed the - sign and though you were confused about the a/2 translation.

See my new post.

 

[assyrian_77]

You can draw contour plots in Mathematica.

 

[mmh37]

Thank you very much! The entire thing does make so much more sense now! :-)

So, is a/2 the focal point then (i.e. where the asymptotes cross)?

 

[quasar987]

a/2 is still a translation. the assymptotes are y=±bx/a

 

[mmh37]

Thanks a lot! It's much appreciated!