Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

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

share|improve this answer
    
Since my parameters are themselves complicated functions, I was hoping to avoid this... –  Felix Goldberg Feb 11 '13 at 16:03
    
Perhaps you should provide this --- and any other relevant information --- in the body of the question. –  Gerry Myerson Feb 11 '13 at 22:30

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:)))

share|improve this answer

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 $y.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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