I got to know that $(x_1 , y_1)$ will be outside the parabola $y^2 = 4ax$ iff ${y_1}^2 > 4ax_1$. I know how to prove it.
My Question Can we say that $(x_1 , y_1)$ will be outside the parabola $Ax^2 + 2Hxy +By^2 + 2Gx +2 Fy +C=0$ $(H^2 = AB)$ iff $A{x_1}^2 + 2H{x_1}{y_1} +B{y_1}^2+ 2G{x_1} +2 F{y_1} +C > 0$ ?
Can anyone please help me to understand?
In short, no. For instance, $f(x,y)=x^2-y=0$ and $g(x,y)=y-x^2=0$ both describe the same parabola, but it should be fairly obvious that for any point $(x_1,y_1)$, $f(x_1,y_1)=-g(x_1,y_1)$.
However, you can salvage this idea by multiplying by a normalizing factor. Let $$f(x,y) = Ax^2+2Hxy+By^2+2Gx+2Fy+C = \begin{bmatrix}x&y&1\end{bmatrix} \begin{bmatrix}A&H&G\\H&B&F\\G&F&C\end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix} = \mathbf x^TQ\mathbf x$$ and $S = \det Q$. For a nondegenerate parabola, $S\ne0$ and the sign of $Sf(x_1,y_1)$ determines whether it’s “inside” or “outside” the parabola represented by $f(x,y)=0$: positive iff inside, negative iff outside, zero if it lies on the curve.