# Level Curves

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.

#### Solution Preview

Hello,

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 ...

#### Solution Summary

This solution is comprised of a detailed explanation to sketch the graph.