Find the Vectorial Equation of the intersection between surfaces $f(x,y) = x^2 + y^2$ and $g(x,y) = xy + 10$

Question

I'm not really sure how to do this, I guessed it had something to do with Vector Functions but overall couldn't find a way to do it. Can you please help?

The equations are:

$$f(x,y) = x^2 + y^2 \ g(x,y) = xy + 10 $$

and I need a Vectorial equation. Thank you in advance!

Answer 1

We look for the intersection of the two surfaces (which are a paraboloid and a hyperbolic paraboloid):

$$\tag{0}\cases{z=x^2+y^2 & (a)\\z=xy+10 & (b)}$$

Let $(r,\theta)$ be the polar coordinates of $(x,y)$; i.e,

$$\tag{1}x=r \cos(\theta), \ \ y=r \sin(\theta).$$

Plugging these expressions in (0)(a) gives

$$\tag{2}r=\sqrt{z}.$$

Using $(0)(b)$, $(1)$ and $(2)$: $z=\sqrt{z} \cos(\theta) \sqrt{z}\sin(\theta)+10$, yielding:

$$z(1-\frac12 \sin(2 \theta))=10 \ \ \iff z=f(\theta) \ \ \text{with} \ \ f(\theta):=\dfrac{20}{2-\sin(2 \theta)}$$

Plugging this expression of $z$ in $(1)$ gives the final description of the intersection curve as a vector function of one variable, the polar angle:

$$\cases{x=\sqrt{f(\theta)}\cos(\theta)\\y=\sqrt{f(\theta)}\sin(\theta)\\z=f(\theta)}$$

(valid for any value of $\theta$ because $f(\theta)>0$ )

Answer 2

$f$ is a paraboloid (red colour), $g$ is a hyperbolic paraboloid (green colour):

(Large versions: link and link)

The intersection reminds me of a Lissajous curve.

For the intersection points we have $$ x^2 + y^2 = xy + 10 \iff \\ 0 = x^2 + y^2 - xy - 10 $$ This seems to be a cylinder with elliptic cross section. $$ x = \frac{1}{\sqrt{2}}(\xi + \eta) \\ y = \frac{1}{\sqrt{2}}(\xi - \eta) $$ So we get \begin{align} 0 &= x^2 + y^2 - xy - 10 \\ &= \frac{1}{2}(\xi^2 + \eta^2 + 2\xi\eta) + \frac{1}{2}(\xi^2 + \eta^2 - 2\xi\eta) - \frac{1}{2}(\xi^2 - \eta^2) - 10 \\ &= \frac{1}{2} \xi^2 + \frac{3}{2} \eta^2 - 10 \iff \\ \end{align} $$ \left( \frac{\xi}{\sqrt{20}} \right)^2 + \left( \frac{\eta}{\sqrt{20/3}} \right)^2 = 1 $$ We can parameterize $$ \xi(t) = \sqrt{20} \cos(2\pi t) \\ \eta(t) = \sqrt{20/3} \sin(2\pi t) $$ for $t \in [0, 2\pi)$ and thus with $z = x^2 + y^2 = \xi^2 + \eta^2$ get \begin{align} x(t) &= \sqrt{10} \cos(2\pi t) + \sqrt{10/3} \sin(2\pi t) \\ y(t) &= \sqrt{10} \cos(2\pi t) - \sqrt{10/3} \sin(2\pi t) \\ z(t) &= 20 \cos(2\pi t)^2 + (20/3) \sin(2\pi t)^2 \\ &= (40/3) \cos(2\pi t)^2 + (20/3) \end{align} The left image shows the ellipse in $\xi$-$\eta$-coordinates (light green), in $x$-$y$-coordinates (pink) and the scene in $3D$ (right image).

(Large versions: link and link)

You can fiddle with an interactive version here:

• adjust the $t$ slider in the left window and see the points on the ellipses ($A(t)$, $B(t)$) and the intersection curve ($C(t)$) change.
• drag the upper part of the right 3D view to change the 3D view