Which normal to $y=x^2$ forms the shortest chord? The line between the pair of points common to the normal and the parabola is the chord formed by that normal.
I tried solving this problem using calculus but I was unable to solve the equation I got from the derivative. Is there a better solution, perhaps using a geometric argument?

My attempt:
I came up with a function for the length of the chord passing through the point $x$ by finding the equation of the normal at $x$, then finding its points of intersection with $y=x^2$ and finally finding the distance between those two points, which came out to be
$$L(x) = \sqrt {\left( 2x + \frac{1}{2x} \right )^2 + \left ( 1 + \frac{1}{4x^2}\right )^2}$$
The minimum of $L(x)$ conincides with the minimum of $L(x)^2$. Hence, I tried to minimize $L(x)^2$. I attempted to solve $\frac{d}{dx} L(x)^2 = 0$ for $x$ however this problem reduced to solving $32x^6 - 6x^2 - 1 = 0$, which I was unable to solve even after substituting $t=x^2$. I plugged the equation into Wolfram to get  $x=\frac{1}{\sqrt{2}}$. So the shortest chord must be the line formed by the points $\left(\frac{1}{\sqrt{2}},\frac{1}{2} \right)$ and $(-\sqrt{2}, 2)$.
 A: Your work is right.
You need only to end it:
For $x=\frac{1}{\sqrt2}$ we get a value $\frac{3\sqrt3}{2}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\left( 2x + \frac{1}{2x} \right )^2 + \left ( 1 + \frac{1}{4x^2}\right )^2\geq\frac{27}{4},$$ which is
$$(2x^2-1)^2(16x^2+1)\geq0.$$ 
An equation of the normal it's:
$$y=-\frac{1}{\sqrt2}x+1$$ for the touching point $\left(\frac{1}{\sqrt2},\frac{1}{2}\right)$ and for $\left(-\frac{1}{\sqrt2},\frac{1}{2}\right)$ it's the similar. 
A: The slope between the points $(a,a^2)$ and $(b,b^2)$ can be calculated two ways,
$$\frac{-1}{2a} = \frac{a^2-b^2}{a-b}$$
The distance squared between the points is
$$(a-b)^2 + (a^2-b^2)^2$$
I rearranged the first formula to plug in to the right part of that second equation since I figured this is easy to plug in and might end up lowering the degree of the polynomial later,
$$-\frac{a-b}{2a} = a^2-b^2$$
$$(a-b)^2 + \frac{(a-b)^2}{4a^2}$$
I then decide to factor,
$$(a-b)^2\left(1+\frac{1}{4a^2} \right)$$
At this point I'm satisfied that I should probably use the first equation to solve for b and plug in and differentiate,
$$\frac{-1}{2a} = a+b$$
$$-b = a + \frac{1}{2a}$$
$$\left( 2a + \frac{1}{2a} \right)^2 \left(1+\frac{1}{4a^2} \right)$$
Clean up the fractions a bit,
$$\frac{(4a^2+1)^2}{4a^2} \frac{(4a^2+1)}{4a^2}$$
$$\frac{(4a^2+1)^3}{16a^4}$$
Now we can differentiate and set this to 0 to obtain our minimum,
$$\frac{3(4a^2+1)^2 4*2a*(16a^4) - (4a^2+1)^3 * 16 * 4 a^3 }{(16a^4)^2} = 0$$
At this point nearly everything divides out,
$$3(4a^2+1)^2 4*2a*(16a^4) - (4a^2+1)^3 * 16 * 4 a^3=0$$
$$6a^2 - (4a^2+1) =0$$
$$2a^2 - 1 =0$$
$$a = \pm \frac{\sqrt{2}}{2}$$
At this point the problem is downhill, but no cubic shows up which seemed to be the problem you were coming to that I never saw.

Furthermore the question is still unanswered because you wanted a purely geometric solution not using calculus. I think that's the real question since it seems possible just like Heron's Problem
A: More of a comment about solving $32x^6 - 6x^2 + 1 = 0$ by hand. After substituting $w=x^2$ you end up with the depressed cubic $32w^3 - 6w - 1 = 0$. At this point you could apply the cubic formula and since everthing is already depressed thats not too hard but before dragging that out don't forget about the rational roots formula. In this case you only have to test +- 1.. 1/32. Starting down the sequence 1/2 works and then you divide it it out to get $16w^2 + 8w + 1$ That's $(4w + 1)^2$ So the other double root is -1/4. However we want the square root of these so the only real root is $\frac{1}{\sqrt{2 }}$
