Hunch on a parabola and a straight line passing through its vertex

Question

Today I came up with a conjecture using GeoGebra software, please, if anyone can help prove it, please do so.

We have a parabola $P$ with focus $F$ and directrix $L$

$M$ is a point on $P$ not on its axis.

Point $M'$ is the projection of $M$ onto the directrix .

Let $A,B$ be the points of intersection of line $M'F$ with $P$.

We draw the tangents $NA,NB$ to $P$.

$N$ is on $L$ because $AB$ is a focal chord.

My conjecture states that $MN$ passes through the vertex of $P$.

Please mention a reference if this feature was previously discovered.

Answer 1

Take the parabola,

$4 p y = x^2$

It's focus $F = (0, p)$ and directrix $y = -p$

Parametric equation of parabola is

$P(t) = ( t , t^2 / 4 p ) $

The tangent direction is given by the derivative of $P(t)$

$P'(t) = ( 1, t / (2 p) )$

Take $t_1$ on the parabola, equation of tangent

$T_1(s) = (t_1 , t_1^2 / 4p ) + s (1 , t_1 / (2p) )$

Take $t_2$ on the parabola, equation of tangent

$T_2(w) = (t_2, t_2^2/ 4 p) + w (1, t_2 / (2p) ) $

If $(1, t_1/2p)$ is perpendicular to $(1, t_2/(2p))$, then

$1 + t_1 t_2 / (4 p^2) = 0$

and from this $t_2 = - 4 p^2 / t_1$

Now, for the intersection of the two tangents, we have, the linear system

$ \begin{bmatrix} 1 && -1 \\ t_1 / (2p) && - t_2 / (2p) \end{bmatrix} \begin{bmatrix} s \\ w \end{bmatrix} = \begin{bmatrix} t_2 - t_1 \\ (t_2^2 - t_1^2) / (4 p ) \end{bmatrix} $

Solution for $s$ is

$s = 1/2 ( t_2 - t_1 )$

Plug this into the expression of $T_1$, the intersection point $N$ is

$N = (t_1 , t_1^2 / (4p) ) + 1/2 (t_2 - t_1) ( 1, t_1 / (2p) )$

$ = ( 1/2 (t_1 + t_2) , 1/(4 p) ( t_1^2 + t_1 t_2 - t_1^2 ) ) $

but $t_1 t_2 / (4 p^2) = -1$

so,

intersection point of the two tangents is $N = ( 1/2 (t_1 + t_2) , - p )$

Now, the intersection of the line passing through $( t_1, t_1^2/(4p) )$ and $(t_2, t_2^2 / (4p) )$ with the directrix is

$L_3: (t_1 , t_1^2/(4p)) + r ( t_2 - t_1, (t_2^2 - t_1^2 ) / (4 p) )$

Set the y-coordinate equal to $-p$, and solve for $r$

$r = (- 4 p^2 - t_1^2 ) / (t_2^2 - t_1^2 )$

the x-ordinate of this intersection point is

$x_{M} = t_1 + (t_2 - t_1) \cdot ( - 4 p^2 - t_1^2 ) / (t_2^2 - t_1^2) = ( - 8 p^2 ) / (t_1 + t_2) $

y-ordinate of $M$ is

$y = 1/(4 p) x^2 = 1/(4 p) . 64 p^4 / (t1 + t2)^2 = 16 p^3 / (t_1 + t_2)^2$

So, now we have the point $N = ( 1/2 (t1 + t2) , - p)$ and $M = ( -8 p^2 / (t1 + t2) , 16 p^3 / (t1 + t2)^2 )$

Connect them with a line and find the x-intercept

$L_4: ( 1/2 (t_1 + t_2) , - p) + w ( - 8 p^2 / (t_1 + t_2) - 1/2 (t_1 + t_2) , p + 16 p^3 / (t_1 + t_2)^2 )$

at $y = 0$

$ p = w ( p + 16 p^3 / (t1 + t2)^2 ) $

so,

$ 1 = w ( 1 + 16 p^2 / (t_1 + t_2)^2 )$

and this gives,

$w = (t_1 + t_2)^2 / ( (t_1 + t_2)^2 + 16 p^2 )$

x-ordinate of this x-intercept point is

$x = 1/2 (t_1 + t_2) + [(t_1 + t_2)^2 / ( (t_1 + t_2)^2 + 16 p^2 )] ( - 8 p^2 / (t_1 + t_2) - 1/2 (t_1 + t_2) )$

simplifying,

$x = 1/2 (t_1 + t_2) + [(t_1 + t_2) / ( (t_1 + t_2)^2 + 16 p^2 )] ( - 8 p^2 - 1/2 (t_1 + t_2)^2 ) = 0 $

Hence, indeed the line joining $M$ and $N$ passes through the vertex $(0,0)$.

Answer 2

without loss of generality we consider $P:4y=x^2$

burnt $F:(0,1)$, $L:y=-1$ vertex $V:(0,0)$.

Let's put $M:(a,a^2/4)$ and therefore $MV:4y=ax$

solving the equations $MV,L$, we find:

$N:(\frac{-4}{a},-1)$

Since $N$ is the result of solving equations $MV$ and $L, M, V, N$ lie on the same line.

We put $M':(a,-1)$ and thus the equation of the line $FM'$ is given by $FM':y-1=\frac{-2}{a}(x)$

Now, from the properties of a parabola, we know that N is the point of intersection of the tangential $P$ in $A,B$ if and only if $AB⊥FN$ is $FM'⊥FN$, let's check this:

$m_{FM'}=\frac{-2}{a}$ , $m_FN=\frac{y_F-y_N}{x_F-x_N}$

$m_{FN}=\frac{1+1}{0+\frac{4}{a}}=\frac{2a}{4}=\frac{a}{2}$

$m_{FM'}×m_{FN}=\frac{-2}{a}×\frac{a}{2}=-1⇒FM'⊥F$

Answer 3

Since you requested a reference, Besant, Conic Sections Treated Geometrically (9th edition, 1895) contains what you want. (download it from the link).

On page 11, Prop X states that tangents at the ends $A,B$ of a focal chord meet at a point $N$ on the directrix, and that $NF$ is perpendicular to $AB$. (The book uses different letters for the points, and you may have to look for earlier proofs. (e.g. Prop VIII)).

On page 23, Prop III states that if line from $M$ through the vertex $A$ intersects the directrix at $E$ then the angle $EFM'$ is a right angle. So $E$ is the same as $N$ (by Prop X above). (If the diagram on pg 23 doesn't quite make sense, see the version on pg 22 with the parabola drawn in.)

So your conjecture is a corollary of these two propositions. But it is not stated explicitly in the book. Which is a bit strange because I'm not sure where else Prop III is used.