Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• Two parabolas have the same point as focus and the same line as axis, but their vertices lie on opposite sides of the focus. Prove that the parabolas intersect orthogonally (i.e., their tangent lines are perpendicular at the points of intersection).
• We know by definition that the vertex is the point midway between the directrix and the focus. Suppose the vertex of one parabola is at $O$ and its focus at $(c, 0),$ then this implies that the second vertex with the same focus is at $(2c, 0).$ If the standard form equation for the first parabola is $y^2 = 4cx,$ then the second parabola, translated by $(2c, 0)$ and which opens in the opposite direction, has standard form equation $y^2 = -4c(x - 2c).$ Taking these equations as functions of y, we get $$x = \frac{y^2}{4c}; \quad x = 2c - \frac{y^2}{4c}$$Setting these equal to one-another, we find the points of intersection occur when $y^2 = 4c^2.$ In other words, the graphs intersect at $(c, -2c)$ and $(c, 2c).$ Taking their respective derivatives gives us $$x' = \frac{y}{2c}; \quad x' = \frac{-y}{2c}$$Evaluated at $(c, 2c)$ we get slopes $1$ and $-1$ respectively, which can be represented by the vectors $(1, 1)$ and $(1, -1),$ the dot product of which is $0.$ Evaluating the derivatives at $(c, -2c)$ yield vectors $(1, -1)$ and $(1, 1),$ whose dot product is also $0.$ Thus, we have proven that the two parabolas intersect orthogonally. $\quad \blacksquare$