How many lines of form $px+2y=1$ are normal to parabola $y^2=4ax$ The line can be written as $$y=-\frac P2 x+1$$
The normal to a parabola is $$y=mx-2am-am^3$$
Comparing them we get 
$$m=-\frac P2$$
And 
$$2am+am^3=-1$$
$$-aP-a\frac{P^3}{8}=-1$$
$$aP+\frac{aP^3}{8}=1$$
Now there is now way to tell. One would think that since it’s a cubic polynomial, we would have 3 values, but the answer is one, so that theory is dismissed. Am I missing something here?
 A: 
Note that the given line $px+2y=1$ passes through the fixed point $(0,\frac12)$, which lies outside the parabola $y^2=4ax$. Geometrically, as seen in the graph, there can be only one normal line to the parabola that passes the point $(0,\frac12)$. (There would be three normals if the fixed point that the line passes lie inside the parabola.)
Analytically, the cubic equation $ap+\frac{ap^3}{8}=1$, which is 
$$p^3+8p-\frac8a=0$$ 
has only one real root because its discriminate is less than zero, i.e.
$$\Delta =-4\cdot8^3 - 27\cdot \frac {8^2}{a^2}\le 0$$
which yields one valid normal as well.
A: Parameterize the parabola as $\gamma(t) = (at^2,2at)$, now notice that the equation of normal at a parameter point $t$:
$$ y+tx= 2at+ at^3$$
And compare this with the original equation $y+ \frac{P}{2} x = \frac12$, we find $t= \frac{P}{2}$ and $ \frac12 = 2at + at^3$, substituting $t$ in terms of $P$ in the cubic:
$$ 0 = aP + \frac{a}{8} P^3 - \frac12$$
Wlog let $a>0$, then we find by Descartes rule of signs that the equation has one root in the interval $(0,\infty)$ and no roots in the interval $(-\infty,0)$
