Find the parabola tangent to $4$ given lines

Question

Given $4$ lines in the Cartesian plane, defined by

$$ a_i x + b_i y + c_i = 0 , \hspace{5pt} i = 1, 2, 3, 4 $$

I want to find the parabola that is tangent to all four lines.

Here is what I did:

Any conic has the general equation

$$ A x^2 + B xy + C y^2 + D x + E y + F = 0 $$

If we define $ r = [x, y, 1]^T $ , then this equation can be put in the form

$$ r^T G r = 0 $$

where

$$ G = \begin{bmatrix} A && \dfrac{B}{2} && \dfrac{D}{2} \\ \dfrac{B}{2} && C && \dfrac{E}{2} \\ \dfrac{D}{2} && \dfrac{E}{2} && F \end{bmatrix} $$

Since this is a parabola then

$$ A C - \left( \dfrac{B}{2} \right)^2 = 0 $$

Now for $i = 1, 2, 3, 4 $ define $N_i = [ a_i, b_i , c_i ]^T $, then it follows that

$$ N_i^T G^{-1} N_i = 0 $$

This equation is linear in the elements of $G^{-1}$. The resulting linear system in the $6$ unknowns and $4$ equations, can be solved to obtain

$$ G^{-1} = \lambda (P_0 + \sigma P_1 ) $$

where $P_0 , P_1 $ are known by solving the system while $\lambda$ and $\sigma$ are arbitrary. We take $ \lambda = 1 $, and note that

$$ G^{-1}_{33} = \dfrac{ \det(G22) }{ \det(G) } $$

where $ G22 $ is the submatrix of $G$ obtained by by deleting the third column and third row.

Now since this is a parabola, as mentioned above, then $ \det(G22) = 0 $, therefore, we want $G^{-1}_{33} = 0 $. Since $G^{-1} = P_0 + \sigma P_1 $, then $\sigma$ is given by

$$ \sigma = - \dfrac{ P_{033} }{ P_{133} } $$

Having obtained $\sigma$, we can calculate $G^{-1}$ and then invert it to obtain $G$.

From $G$, we now have all the coefficients of its general equation up to a scale. Now it is possible after some manipulation of these coefficients to obtain the vertex and the orientation of the parabola, and, of course, its squared term multiplier. This completes the specification of the desired parabola.

I would greatly appreciate your feedback on this method that I developed, and also other solutions that you might have to the problem.

Answer 1

Here's a very simple geometric construction to find the parabola tangent to four given lines.

The construction relies on a beautiful property of parabolas:

The focus of a parabola lies on the circumcircle of the triangle formed by any three tangents to the parabola (see Appendix for a geometric proof).

Let then $a$, $b$, $c$, $d$ be four given lines (no two of them parallel). The focus $F$ of the parabola tangent to them must lie (by the above property) on the circumcircle of the triangle formed by lines $abc$, and at the same time it must lie on the circumcircle of the triangle formed by lines $abd$. Those two circles have in common the intersection point of lines $a$ and $b$, which cannot be the focus. Hence the focus of the parabola is uniquely determined as the other intersection of the circumcircles.

To complete the construction we can then use another well-known property:

The perpendicular projection of the focus of a parabola onto any tangent, lies on the line passing through the vertex and perpendicular to the axis of the parabola.

If $A$ and $B$ are the projections of the focus on lines $a$ and $b$, the line through the focus perpendicular to line $AB$ is then the axis, and its intersection with $AB$ is the vertex of the parabola.

APPENDIX.

Proof of the property given at the beginning.

Let three tangents to a parabola be given, with tangency points $A$, $B$, $C$ and intersecting at $P$, $Q$, $R$ (see figure below). Let $F$ be the focus of the parabola, and $D$, $E$ the intersections of the axis of the parabola with lines $PQ$ and $QR$.

It is well known that $FA=FD$, which entails $\angle FAQ=\angle EDQ$. In addition, we know that (see here for a proof):

the exterior angle between any two tangents is equal to the angle which either segment of tangent subtends at the focus,

which in our case implies $\angle AFQ=\angle DQE$.

It follows that in triangles $AFQ$, $DEQ$ we also have $\angle FQA=\angle DEQ$.

But with an analogous reasoning we can prove that $\angle FRB=\angle FER=\angle DEQ$. It follows that $\angle FRP=\angle FQP$, and points $FPQR$ are concyclic as it was to be proved.