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You are given a second order surface $Ax^2+By^2+Cz^2=α$ and a plane $x+z=β$. Determine the type of the surface and the type of the curve at the intersection of the surface and the plane.

Basically, I need to write a program, that will do task, getting as parameters $A, B, C, \alpha,\beta$. So, the best way I found to make this, is to just check every equation by coefficients signs to determine quadratic surface. But what about curves of intersection? Is there a simple way to determine their type only by given parameters?

Prox
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3 Answers3

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One of the ways would be to parametrize your plane and all points on the plane can be then represented as $(t, y, \beta - t)$.

Plugging it into the equation of your quadratic surface, you get parametrized equation of the intersection curve.

$r(t,y) = At^2+By^2+C(\beta - t)^2 - α = 0$

i.e $ \,r(t,y) = (A+C)t^2+By^2 -2 C \beta \,t - α + \beta^2C = 0$

Math Lover
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Comment:

When you intersect the surface with plane $p(x, z)=x+z=\beta$ that means you eliminate y in the surface equation. Now you have curve $Ax^2+Cz^2=\alpha$ in (x,z) plane; we may rewrite this equation as:

$\frac {x^2}{C}+\frac{z^2}{A}=\frac {\alpha}{AC}$

The coordinate of points of intersection are the solution of this system of equatn:

$\begin{cases}\frac {x^2}{C}+\frac{z^2}{A}=\frac {\alpha}{AC}\\x+z=\beta \end{cases}$

If $A, C. \alpha>0$ then you have an ellipse. If [$A<0$ and $C>0$]or [$C<0 $ and $A>0$], then Hyperbola is the intersection.

sirous
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  • We cannot assume $y$ to be zero and write $Ax^2+Cz^2=\alpha$. The reality is that all values of $y$ are on the given plane. The given quadratic puts a constraint on which values of $y$ would be on the intersection curve. The intersection curve is not in the $XZ$ plane. – Math Lover Dec 05 '20 at 12:41
  • @MathLover, plane $z+x=\beta$ is parallel to y axis. Of course any point on this plane has also y coordinate. $x+z=\beta$ is a lone in xz plane but in 3D it is a plane parallel to y axis. that means for any value of y relation $x+z=\beta$ holds. – sirous Dec 05 '20 at 12:51
  • Yes but $Ax^2+Cz^2=\alpha$ is only true for $y = 0$, What about points on the intersection curve where $y \ne 0$? – Math Lover Dec 05 '20 at 12:55
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Regarding the existence of intersections. Making the substitution of $x = \beta-z$ into $A x^2+B y^2+C z^2=\alpha$ we get

$$ B y^2 + C z^2+(\beta-z)^2 = \alpha $$

so

$$ y = \pm\frac{\sqrt{\alpha-A (\beta -z)^2-C z^2}}{\sqrt{B}} $$

and the intersection exists if $\Delta = \alpha-A (\beta -z)^2-C z^2\ge 0$ assuming $B\gt 0$

Now

$$ \Delta = \frac{(A+C) \alpha -A C \beta ^2}{A+C}-(A+C) \left(z-\frac{A \beta }{A+C}\right)^2 \ge 0 $$

and assuming for instance that $A+C>0$ if

$$ (A+C) \alpha \lt A C \beta ^2 $$

no intersection.

Cesareo
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