Animated Solution for Mathematics - Conic Sections: What normal to the curve $y=x^2$ forms the shortest chord?

Imagine you are standing on the graph of the beautiful, symmetric parabola defined by $y=x^2$. It is the most fundamental curve in physics, representing the path of a projectile or the shape of a satellite dish.

Today, we are going to explore a hidden property of this curve: the shortest normal chord. A normal chord is a line segment that is perpendicular to the tangent at one point on the curve and intersects the curve at another point. We want to find which of these chords is the shortest.

The Parametric Setup

To make our math elegant, we avoid the clunky Cartesian coordinates $(x,y)$ where possible. Instead, we use the parametric form. We pick a point $P$ on the curve and define it as $(t,t^2)$.

This is a powerful choice because it automatically satisfies the equation $y=x^2$. Now, we need the slope of the tangent at $P$. Differentiating $y=x^2$ gives us $\frac{dy}{dx}=2x$.

At point $P$, the slope of the tangent is $2t$. Since the normal is perpendicular to the tangent, its slope $m_n$ must be:

The Equation of the Normal

With the slope $m_n=-\frac{1}{2t}$ and the point $P(t,t^2)$, we use the point-slope form:

If we multiply both sides by $2t$ and rearrange, we get the elegant equation of the normal:

This line represents every possible normal at point $P$. But this line doesn't just exist in isolation; it slices through the parabola at another point, $Q$. Let $Q$ have the coordinates $(t_1,t_1^2)$.

Finding the Intersection Point $Q$

Since $Q$ lies on the normal line, its coordinates must satisfy the normal equation. Substituting $x=t_1$ and $y=t_1^2$ into $x+2ty=t+2t^3$, we get:

This looks like a mess, but let's bring everything to one side:

We can factor out $(t_1-t)$ to get $(t_1-t)[2t(t_1+t)+1]=0$. Since $P$ and $Q$ are distinct, $t_1 eq t$, so we can divide by $(t_1-t)$. This leaves us with the crucial relationship:

The Length of the Chord

Now, we define the length $L$ of the chord $PQ$. Using the distance formula, $L^2=(t-t_1{)}^2+(t^2-t_1^2{)}^2$. Factoring out $(t-t_1{)}^2$, we get:

Substituting our expression for $t_1$, we find that $(t-t_1)=2t+\frac{1}{2t}$ and $(t+t_1)=-\frac{1}{2t}$. Plugging these into our $L^2$ expression, we get:

Simplifying this, we arrive at the compact form:

The Final Optimization

We are at the finish line! To minimize $L^2$, we differentiate with respect to $t$ and set it to zero. To make this easier, we use the substitution $u=t^2$.

Our function becomes:

Applying the quotient rule and setting the numerator to zero, we find $2u-1=0$, which means $u=\frac{1}{2}$. Since $u=t^2$, we have $t=\pm \frac{1}{\sqrt{2}}$.

Substituting these values back into our normal equation, we find the two lines:

These are the normals that form the shortest chords. You have successfully navigated the geometry and the calculus to find the answer. Well done!