Sketch the graph:
z= f(x,y) where f(x,y)= x^2 + xy ; c=0,1,2,3,-1,-2,-3
Please show/describe all steps AND explain why it is a hyperbola.
Also, sketch the graph:
z=f(x,y) where f(x,y)= x/y ; c= 0,1,2,3,-1,-2,-3
Please show/describe all steps.
I have attached the explanation below as a txt file which should make the 2x2 matrices easier to read. A sketch of the hyperbola is also attached as a jpeg file - if opened in Internet Explorer please maximise (by clicking in lower right corner of image).
I have tried to use your equation to give a general method for recognizing conic sections, so you can use the same method to sketch similar looking equations in the future, so apologies if it is a little long winded for the given question.
We have the function z = f(x,y) = x^2 + xy, so the level curve at height z = c
is given by:
x^2 + xy - c = 0 (1)
This is an equation of the form:
A x^2 + B xy + c y^2 + F x + G y + H = 0 (2)
(the general form for a conic section)
with A=1, B=1, C=F=G=0, and H=-c
We know the standard form for an hyperbola is:
x^2/(a^2) - y^2/(b^2) = 1 (3)
and that this standard form would tell us useful properties of the hyperbola in terms of
a and b:
i) There is a directrix at y = (a/b)x and also at y = -(a/b)x
(These are the asymptotes to which the two branches of the hyperbola tend towards.)
ii) The hyperbola cuts the x axis at x = a
Both of which we can see directly from the equation in standard form.
If we take a hyperbola in its standard form (3) and imagine rotating about the origin
through angle theta, and then apply a translation by (x,y), you can see we still have an
hyperbola of exactly the same shape - just at a different place and orientation in the xy
plane. After the transformations, its equation ...
This solution is comprised of a detailed explanation to sketch the graph.