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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?

Quanto
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Aditya
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  • What is you definition of a normal line to the parabola? That the line is orthogonal to the tangent line at some point, and is going through that same point? Is $a$ given, so the parabola fixed, or is $p$ given and the line fixed, or is it about all combinations? – Henno Brandsma Feb 25 '20 at 13:18
  • @HennoBrandsma $a$ is not given – Aditya Feb 25 '20 at 14:17

2 Answers2

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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 $\Delta =-8^2 (4+\frac {27}{a^2})\le 0$, yielding one valid normal as well.

Quanto
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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 $(-\infty,0)$

tryst with freedom
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