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

Question

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?

Answer 1

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..

Answer 2

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 $f(p,q)\big\{^<_>\big\}0$.