I am stuck on what seems to be an easy exercise.
We have $f(x,y) = x^2 + 4xy + y^2 \mbox{ for all } (x,y)$ in $\mathbb{R}^2.$
Now we are supposed to find a parametrization of the intersection curve between $f(x,y)$ and $z = x + 3y.$
I've been stuck for hours now, anyone got any ideas/tips?
Thank you!
$\endgroup$ 14 Answers
$\begingroup$I'm not sure that this is an 'easy' exercise, exactly, at least not at the level one would encounter such a problem, but the difficulties are more computational than conceptual.
We want to build a parameterization of the set of intersection of two surfaces given as graphs of functions $f, L$. By construction, the solution set of the equation $$f(x, y) = L(x, y)$$ is the projection of the set of intersection to the $xy$ plane. So, if we can find a parameterization $t \mapsto (x(t), y(t))$ of that curve, then the desired parameterization of the intersection of the graphs is just the image of that curve under either function, namely,$$t \mapsto (x(t), y(t), L(x(t), y(t))) .$$
In our situation, as is often the case, it's easier to make a change of coordinates in which the expressions for the surfaces are easier to work with; once we've parameterized the intersection in the new coordinates, if we'd like we can produce a parameterization in the given coordinates simply by reversing the changes of variable.
If we regard $f$ as a quadratic form, its discriminant is $4^2 - 4(1)(1) = 12 > 0$, so there is a linear change of coordinates $(x, y) \rightsquigarrow (u, v)$ in which $f$ is given by$$f(u, v) = u^2 - v^2 .$$
It's a standard algebra exercise to find such a change of coordinates, and we can simplify the process by observing that $f$ is invariant under the reflection in the lines $y = \pm x$.
One linear transformation that does this is $x = \tfrac{1}{\sqrt{6}} u + \tfrac{1}{\sqrt{2}} v$, $y = \tfrac{1}{\sqrt{6}} u - \tfrac{1}{\sqrt{2}} v$.
The linear function $L : (x, y) \mapsto x + 3 y$ in the new coordinates (since the change of coordinates is linear) has the form $(u, v) \mapsto 2 a u + 2 b v$.
Now, setting $f$ and $L$ equal gives$$u^2 - v^2 = 2 a u + 2 b v,$$and rearranging and completing the square gives$$(u - a)^2 - (v + b)^2 = a^2 - b^2 .$$It turns out that $a^2 - b^2 \neq 0$, so by making another suitable linear change of variable, we can write this equation as$$r^2 - s^2 = 1 ,$$which is the equation for the usual unit hyperbola, and which can be parameterized by $$t \mapsto (\pm \cosh t, \sinh t).$$Again in $rs$-coordinates the plane is given by the equation $(r, s) \mapsto c r + d s$, so substituting gives that the intersection curve (which has two components) is parameterized by$$t \mapsto (\pm \cosh t, \sinh t, \pm c \cosh t + d \sinh t) .$$
If we prefer a rational parameterization, we can substitute $t = 2 \operatorname{artanh} p$, so that $\cosh t = \frac{1 + p^2}{1 - p^2}$, $\sinh t = \frac{2 p}{1 - p^2}$.
$\endgroup$ $\begingroup$Hint: First you find the set of points that intersect.Solve f(x,y)=z. Then try parametrizing the curve you get have in mind known curves as circles-parabolas-lines or planes or hyperbola most likely will be a hyperbola Can you see why?Your first equation is a hyperbolic paraboloid which is easy to see .What happens if you cut it with a plane like x+3y?
$\endgroup$ 2 $\begingroup$$$ x^2 +4xy +y^2 -x -3y = (x+2y)^2 -3y^2 -(x+2y) -y = (X-1/2)^2 -3(Y-1/6)^2 -17/36=0 $$ so the projection of intersection forms a hyperbola.
$\endgroup$ $\begingroup$Setting 'z= z' gives $x^2+ 4xy+ y^2= x+ 3y$. The first thing I would do is "complete the square" on the left: $x^2+ 4xy+ 4y^2- 3y^2= (x+ 2y)- 3y^2= x+ 3y$. Now, let u= x+ y and v= y then x+ 3y= u- v+ 3v= u+ 2v so the equation becomes $u^2- 3v^2= u+ 2v$ or $u^2- u- 3(v^2+ (2/3)v)= 0$ and complete the square again: $u^2- u+ 1/4- 3(v^2+ (2/3)v+ 1/9)= (u- 1/2)^2- 3(v+ 1/3)^2= 1/4+ 1/3= 7/12$. Let p= u- 1/2 and q= v+ 1/3 so the equation becomes $p^2- 3q^2= 7/12$ or $\frac{p^2}{\frac{12}{7}}- \frac{q^2}{\frac{4}{7}}= 1$
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