Prove that the parabloas are mutually perpendicular.

Question

Given that two parabolas have the same focus with their axes of symmetry in opposite directions. Then I have to prove that the two intersect at right angles. As I think, because the axes are oppositely directed, the directrices must be rotated by an angle $π$ . So let their equations be $x+y+c=0$ and $x-y+c=0$. Let the common focus of the two parabolas be, for simplicity, the origin $(0,0)$. Then the two parabolas are: $$F(x,y):=x^2+y^2-(x+y+c)^2=0$$ $$G(x,y):=x^2+y^2-(x-y+c)^2=0$$ Hence , the angle between the two parabolas comes out to be: $$atan \frac {F_xG_y - F_yG_x}{F_xG_x + F_yG_y}$$ $$= atan \frac {2c(c+x)}{x^2+2cx+y^2}$$ Where am I getting wrong? The subscripts stand for partial differentials.

Answer 1

If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are: $$\Gamma_\pm:=(x^2+y^2=(y\mp c)^2)=(x^2\pm2yc-c^2=0)$$ They intersect at $(\pm c,0)$ and it is easy to see their tangents have slope $\pm 1$ there.

If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives $$ \Gamma_i:=(x^2+2yc_i-c_i^2=0) $$ so $y=\frac12(c_1+c_2)$ and $x=\pm\sqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-\frac{c_i}{x}$ and so multiplies to $-1$.