Parabolas and Loci

Question

There's a parabola $y = 4x^2$. The point P moves along the curve. A line passing through P and the point $(0, 1)$ intersects the curve again at Q. The tangents to the curve at P and Q intersect at X. Describe the locus of X as P moves along the curve.

Here's a Desmos visual in case it helps: https://www.desmos.com/calculator/d9davklusv

My initial thoughts were as the x coordinate of P gets very large, Q will approach the origin. Thus for very large P, the point X will be close to the x axis (since the tangent at Q will be almost a horizontal line and P will be almost a vertical line), however after playing around on Desmos, this is incorrect.

Does anyone have suggestions on how to come up with the locus without using excessive algebra, many thanks in advance :)

Answer 1

Consider the parabola $ 4 p y = x^2 $ , with $ p \gt 0 $ is the focal length.

In compact matrix-vector notation, the equation is

$ r^T Q r + b^T r = 0 $

where

$ Q = \begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix}, \hspace{40pt} b = \begin{bmatrix} 0 \\ - 4 p \end{bmatrix} , \hspace{40pt} r = \begin{bmatrix} x \\ y \end{bmatrix}$

Now consider a point $S = (x_0, y_0)$ from which you drop tangent to the parabola. The normal vector to the parabola as function of $r$ is given by

$ g = 2 Q r + b $

And from tangency, we want

$ g \cdot (r - S) = 0 $

That is,

$ (2 Q r + b) \cdot (r - S) = 0 $

Since $ u \cdot v = u^T v $, then this last equation becomes

$ 2 r^T Q r - 2 r^T Q S + b^T r - b^T S = 0 $

Using the fact that $ r^T Q r + b^T r = 0 $, this simplifies to

$ - b^T r - 2 r^T Q S - b^T S = 0 $

Substitute all the variables to write this last equation in terms of $x, y$:

$ - 2 x x_0 + 4 p y + 4 p y_0 = 0 $

This shows that if $(x, y)$ is a tangent point then it lies on this line. Substitute $x = 0$ , you get $ y = - y_0 $, i.e. the line segment between the two tangency points passes through $(0, -y_0)$

Conversely, if we have any line segment on the parabola, that passes (or its extension) through $(0, y_1)$, and we draw the tangents at the end points then they must meet a point whose $ y $ coordinate is $ - y_1 $.

Therefore, since $PQ$ in the above problem pass through $(0,1)$, the locus of the intersection point of the tangents will be the line $y = -1$.

Answer 2

Consider the parabola $a'$ having $F=(0,1)$ as focus and $y=-1$ as directrix, whose equation is $y={1\over4}x^2$. This parabola is obtained by applying to the parabola $a$ of equation $y=4x^2$ the transformation $$ x\to{4}x,\quad y\to y. $$ Under that transformation points $P$, $Q$ are mapped to $P'$, $Q'$, the tangents to $a$ at $P$, $Q$ and their intersection $X$ are respectively mapped to the tangents to $a'$ at $P'$, $Q'$ and their intersection $X'$. Note that line $P'Q'$ passes through $F$, because $F$ is mapped to itself.

But we know that the locus of $X'$ is the directrix $y=-1$ of $a'$, because the lines tangent to a parabola at the endpoints of a focal chord are perpendicular and meet on the directrix. The locus of $X$ is then the inverse image of that line, which is the same line $y=-1$.