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 $(-\sqrt{2}, 2)$.

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
