[Math] How to construct the point of intersection of a line and a parabola whose focus and directrix are known

conic sectionsgeometric-construction

I found this problem in Polya's "How to solve it". It goes as follows

Using only a straight edge and a compass, construct the point(s) of intersection of a given line and a parabola whose focus and directrix are known.

I think I've managed to do it for the special case when the directrix and the given line are parallel, but I'm stuck at the general case when they're not. I could add my method for the special case in my question but I think it's convoluted and inelegant. I was wondering whether anyone could help me figure it out. I could just look it up in the book, but Polya just uses that problem to illustrate restatement of a problem and does not provide a solution. I'd be glad if someone could give me some pointers.

Best Answer

Using the definition of parabola, each of the intersection points $P_1,P_2$ lies on L and is the center of a (different) circle through point F and tangent to D (e.g. $|P_1F| = |P_1T|$). But this only gives two pieces of information for each circle; three are needed.

See diagram.

The Original Problem

But, the missing third point can be obtained by reflecting $F$ about the line $L$ to get $F'$ ($P_1$ and $P_2$ lie along L and so their circles must be symmetrical about L). See diagram.

Finding Another Point of Interest

Then extend $FF'$ until it intersects $D$ at $Q_1$, and intersects with $L$ at point $C$. Now the power of $Q_1$ relative to the circle centered at $P_1$ can be expressed as

$|Q_1T_1|^2 = |Q_1F||Q_1F'|$

Draw semi-circles with diameter $FF'$ and diameter $Q_1C$. These intersect at point $R_1$ such that

  • $\angle QR_1C = 90^\circ$ and
  • $|Q_1R_1|^2 = |Q_1F||Q_1F'|$.

Therefore $|Q_1R_1| = |Q_1T|$

So a circular arc centered at $Q_1$ through point $R_1$ will intersect $D$ at the desired tangency point $T_1$. Just draw a perpendicular line through $T_1$ to find $P_1$.

To find $P_2$ extend the semi-circles to find another point of intersection $R_2$ (not shown) which will lie on the same arc as $R_1$ and $T$.

So just extend the arc centered at $Q_1$ until it intersects the diameter again at $T_2$. A perpendicular bisector yields point $P_2$.

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