2
$\begingroup$

Suppose we have a second-order curve in general form:

(1) $a_{11}x^{2}+2a_{12}xy+a_{22}y^{2}+2a_{13}x+2a_{23}y+a_{33}=0$.

I'd like to know if there is a simple condition that ensures that the curve has at least one point on on or below the $x$ axis, i.e. that the left-hand side of (1) is nonpositive.

In the trivial case that the curve is a parabola, the discriminant being nonnegative is just such a condition. But what happens in the general case?

$\endgroup$

3 Answers 3

2
$\begingroup$

Solve for $y$ in the form $y= A(x) \pm \sqrt{B(x)}$ and estimate. More abstract versions are just variant of this.

$\endgroup$
2
  • $\begingroup$ Since my parameters are themselves complicated functions, I was hoping to avoid this... $\endgroup$ Feb 11, 2013 at 16:03
  • $\begingroup$ Perhaps you should provide this --- and any other relevant information --- in the body of the question. $\endgroup$ Feb 11, 2013 at 22:30
1
$\begingroup$

We may regard the left-hand side of the equation of thecurve as a quadratic polynomial in $x$. If $D(y)$ is its discriminant (with respect to $x$), then $D(y)\ge 0$ iff there exists a point with the second coordinate $y$ on the curve. Solve this inequality for $y$ and check whether its minimal solution is negative:)))

$\endgroup$
0
$\begingroup$

I am not sure I understand the question. A point ON the $x$ axis is given by $y=0,$ so $a_{11} x^2 + 2 a_{13} x + a_{33}=0.$ That IS a parabola, and so we know what the solutions are (if any). If if the discriminant is positive, you are golden. If the discriminant is 0, pick a random value of $x$ ($0$ is easiest, but that might be the solution to your quadric), and check if $y$ is positive. If the discriminant is negative, take $x = 0$ and see what the solutions are to the resulting quadric in $y.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.