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What is the condition of the point $(x_1 , y_1)$ being outside the parabola $Ax^2 + 2Hxy +By^2 + 2Gx +2 Fy +C=0$ $(H^2 = AB)$?

I am totally clueless how to prove that...

Can anyone please help me?

cmi
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2 Answers2

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Let's suppose the axis of the parabola as $X'$ axis and the line which passes through the vertex of the parabola and parallel to the directrix as $Y'$ axis. Then you will get to see that ${y'_1}^2 > 4a{x'_1}$ where $y'_1$ and $x'_1$ are the distance of $(x_1 , y_1)$ from the $X'$ and $Y'$ axes respectively.

Now to get the original inequation put $y'_1 =x_1 Sin M + y_1 Cos M+k$ and $x'_1 =x_1 Cos M - y_1 Sin M + h$ . Where $M$ is the rotational angle to be rotated to get the original axes and $(h ,k)$ is the co ordinate of the original origin with respect to $X'$, $Y'$.

Thus you will get the given condition for the point being outside of the parabola..

anonymous
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Let $f(x,y)=Ax^2 + 2Hxy +By^2 + 2Gx +2 Fy +C$, where both $A,B\ge0$.

The point $(p,q)$ is $\big\{^\text{inside}_\text{outside}\big\}$ the parabola if $f(p,q)\big\{^<_>\big\}0$.

Hypergeometricx
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  • This test fails for $f(x,y)=y-x^2$. – amd Sep 04 '19 at 19:38
  • Thanks. Have now specified that coefficients of the squared terms should be positive. – Hypergeometricx Sep 05 '19 at 13:56
  • Umm... that might be a bit much: $x^2-y$ doesn’t meet this criterion. I usually recommend multiplying by the determinant of the associated $3\times3$ matrix to normalize, but maybe neither $A\lt0$ nor $B\lt0$ is sufficient. Or, the quadratic part can be rewritten as $\pm(\alpha x+\beta y)^2$, and it’s this term that needs to be nonnegative. – amd Sep 05 '19 at 17:54
  • Noted. Amended to say $A,B\ge 0$. – Hypergeometricx Sep 06 '19 at 14:27