# Can we say $(x_1 , y_1)$ will be outside the parabola iff $A{x_1}^2 + 2H{x_1}{y_1} +B{y_1}^2+ 2G{x_1} +2 F{y_1} +C > 0$?

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 $$A{x_1}^2 + 2H{x_1}{y_1} +B{y_1}^2+ 2G{x_1} +2 F{y_1} +C > 0$$ ?

• Yes , it will. Another conic section is $xy=c$.you can see it's true.You can also use transformation – Rishi Sep 4 at 12:55
• I don't understand this question. If the parabola is tilted with respect to the axes, its equation will of course not be $y^2 = 4ax$, so you can't expect to gain any information by plugging coordinates into that equation. And what do you mean by an equation giving a negative value? – Hans Lundmark Sep 4 at 13:42
• I do not want to plug $(x_1 , y_1)$ in $y^2 = 4ax$.. I would plug $(x_1 , y_1)$ in the equation of the parabola whose axis is not parallel to any of the axes..I want to know if the condition for point being outside the parabola remains same or not?@Hans Lundmark – cmi Sep 4 at 14:12
• I have edited my question..Can you please check ?@HansLundmark – cmi Sep 4 at 15:33
• How ? Can you explain?@Rishi – cmi Sep 4 at 15:33

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.