# Thread: Line Tangent To Parabola's

1. ## Line Tangent To Parabola's

Just Needed Help on this:
A line is Tangent to a parabola $Y^2=4ax$ at a point P(not the Vertex) and intersects the axis of this parabola at a point Q. Prove that the line segment PQ is bisected by the line which is tangent to the parabola at its vertex.

2. Hello, Rimas!

A line is tangent to a parabola $y^2=4ax$ at a point $P$ (not the vertex)
and intersects the axis of this parabola at a point $Q.$
Prove that the line segment $PQ$ is bisected by the line
which is tangent to the parabola at its vertex.

If they made a sketch, they could have stated the problem more clearly.
Code:
                |
|      P   o     *
|      o *
|  o* (p,q)
o| *
o    |*
o        |
-o- - - - - - * - - - - - - - - -
Q            |
|*
| *
|   *
|
The parabola is "horizontal", vertex at (0,0), opens to the right.
The tangent at $P(p,q)$ has its x-intercept at $Q.$
Prove that $PQ$ is bisected by the y-axis.

We have: . $y^2 = 4ax \quad\Rightarrow\quad 2yy' = 4a \quad\Rightarrow\quad y' = \frac{2a}{y}$

At $P(p,q)$, the slope of the tangent is: . $m = \frac{2a}{q}$

The equation of the tangent is: . $y - q \:=\:\frac{2a}{q}(x-p) \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{q^2-2ap}{q}$

Since $q^2 = 4ap$, we have: . $y \:=\:\frac{2a}{y}x + \frac{4ap - 2ap}{q} \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{2ap}{q}$

Let $y = 0$, the x-intercept is: . $Q(-p,0)$

Let $x = 0$, the y-intercept is: . $M\left(0, \frac{2ap}{q}\right)$ .[1]

Since $q^2 = 4ap$, then $2ap = \frac{q^2}{2}$

. . Substitute into [1]: . $M\left(0, \frac{\frac{q^2}{2}}{q}\right) \:=\:M\left(0,\frac{q}{2}\right)$

And $M\left(0,\frac{q}{2}\right)$ is indeed the midpoint of $P(p,q)$ and $Q(-p,0)$