= Solution
The two real slope roots of $p^2+fp-1=0$ are distinct because their discriminant is $f^2+4>0$. <Vieta's formulas> give $p_+p_-=-1$, so the tangent vectors $(1,p_+)$ and $(1,p_-)$ have zero <scalar product>. Whenever the coefficients define local solution curves, the two families are therefore <orthogonal trajectories>.
For the stated example at $(3,4)$, the slopes are $3$ and $-1/3$. In the family from part <a>, the constants are $1$ and $-1/9$, giving
$$
\boxed{y=\frac{x^2-1}{2},\qquad y=\frac92-\frac{x^2}{18}.}
$$
Both curves pass through $(3,4)$, and $3(-1/3)=-1$ verifies the right angle.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2021-orthogonal-parabolas.png]
{title=Two members of the same parabolic family pass through (3,4) with perpendicular tangent lines}
Back to article page
