Lines $p$, $q$ and $d$ are given. Construct a parabola tangent to $p$ and $q$ so that $d$ is its directrix.
I don't even know how to approach this problem. I would appreciate any ideas.
Disclaimer: No analytic geometry allowed solving this problem
Lines $p$, $q$ and $d$ are given. Construct a parabola tangent to $p$ and $q$ so that $d$ is its directrix.
I don't even know how to approach this problem. I would appreciate any ideas.
Disclaimer: No analytic geometry allowed solving this problem
On
Let the parabola $C$ be given by its focus/directrix equation: $$\begin{align}C: (x-x(F))^2+(y-y(F))^2-(a_d x+b_d y+ c_d)^2/(a_d^2+b_d^2)&=0\\ p: a_p x+b_p y+ c_p&=0\\ q: a_q x+b_q y+ c_q&=0\end{align}$$
and evaluate the dual conic of $C$ at $(\frac{a_p}{c_p},\frac{a_p}{c_p})$ and $(\frac{a_q}{c_q},\frac{a_q}{c_q})$ which gives you two equations in $x(F),y(F),$ with solution:
$$(x(F),y(F))=\\((((b_d^2b_p^2+2a_da_pb_db_p-a_p^2b_d^2)b_q+a_da_qb_db_p^2+2a_d^2a_pa_qb_p-a_da_p^2a_qb_d)c_q+(((-b_d^2b_p)-a_da_pb_d)b_q^2+((-2a_da_qb_db_p)-2a_d^2a_pa_q)b_q+a_q^2b_d^2b_p+a_da_pa_q^2b_d)c_p+((a_p^2b_d-a_da_pb_p)b_q^2+(a_da_qb_p^2+a_da_p^2a_q)b_q-a_q^2b_db_p^2-a_da_pa_q^2b_p)c_d)/((a_pb_d^2+a_d^2a_p)b_pb_q^2+(((-a_qb_d^2)-a_d^2a_q)b_p^2+a_p^2a_qb_d^2+a_d^2a_p^2a_q)b_q+((-a_pa_q^2b_d^2)-a_d^2a_pa_q^2)b_p),(((a_db_db_p^2-2a_pb_d^2b_p-a_da_p^2b_d)b_q+a_d^2a_qb_p^2-2a_da_pa_qb_db_p-a_d^2a_p^2a_q)c_q+(((-a_db_db_p)-a_d^2a_p)b_q^2+(2a_qb_d^2b_p+2a_da_pa_qb_d)b_q+a_da_q^2b_db_p+a_d^2a_pa_q^2)c_p+((a_pb_db_p+a_da_p^2)b_q^2+((-a_qb_db_p^2)-a_p^2a_qb_d)b_q-a_da_q^2b_p^2+a_pa_q^2b_db_p)c_d)/((a_pb_d^2+a_d^2a_p)b_pb_q^2+(((-a_qb_d^2)-a_d^2a_q)b_p^2+a_p^2a_qb_d^2+a_d^2a_p^2a_q)b_q+((-a_pa_q^2b_d^2)-a_d^2a_pa_q^2)b_p))$$

The reflection of the directrix about any tangent passes through the focus of the parabola (see proof below).
This suggest an easy way to solve your problem: construct lines $p'$ and $q'$, reflections of $d$ about $p$ and $q$. The focus of the parabola is the intersection between $p'$ and $q'$.
PROOF. Let $P$ be a point on a parabola, $T$ the point where the tangent at $P$ intersects the directrix and $H$ the projection of $P$ on the directrix. If $F$ is the focus, then the tangent is the bisector of $\angle FPH$ and $PF=PH$. Hence triangles $FPT$ and $HPT$ are congruent and line $TF$ is the reflection of the directrix about the tangent.