Animated Solution for Mathematics - Differentiation: Let $P$ and $Q$ be any points on the curves $(x-1{)}^2+(y+1{)}^2=1$ and $y=x^2$, respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval

Imagine you are standing on a vast, flat plane. In front of you, there are two distinct shapes: a circle defined by the equation $(x-1{)}^2+(y+1{)}^2=1$ and a parabola defined by $y=x^2$.

Your mission is to find the absolute shortest distance between any point $P$ on the circle and any point $Q$ on the parabola. In the world of coordinate geometry, there is a beautiful, hidden symmetry that makes this problem collapse into a simple, elegant solution.

The Common Normal

Before we dive into the algebra, let's pause and visualize. The shortest distance between two non-intersecting curves is never a random line; it is always the segment of the common normal.

Think of it as a bridge that is perfectly perpendicular to both curves at the points of contact. For our circle, any normal line is special—it is guaranteed to pass through the center $C(1,-1)$.

If the shortest distance lies along a line that is normal to both, then that line must pass through the center of the circle. Therefore, the normal to the parabola at point $Q$ must also pass through $C(1,-1)$. This realization is the key that unlocks the entire problem.

Let's focus on the parabola $y=x^2$. We can represent any point $Q$ on this curve as $(t,t^2)$.

To find the normal at this point, we first need the slope of the tangent. Using basic calculus, we differentiate $y=x^2$ to get:

At our point $Q$, the slope of the tangent is $2t$. Since the normal is perpendicular to the tangent, its slope $m_n$ is the negative reciprocal:

Now, we can write the equation of this normal line using the point-slope form:

We know this normal line must pass through the center of the circle, $C(1,-1)$. This is our constraint. By substituting $x=1$ and $y=-1$ into our normal equation, we get:

To solve for $t$, we multiply both sides by $2t$ to clear the denominator:

Rearranging the terms, we arrive at a beautiful cubic equation:

We don't need to solve this cubic equation exactly. We just need to know where its root lies to match it with our requirements. Let $f(t)=2t^3+3t-1$.

By testing the boundaries of our given intervals, specifically $t=\frac{1}{4}$ and $t=\frac{1}{2}$, we find that $f(\frac{1}{4})<0$ and $f(\frac{1}{2})>0$. By the Intermediate Value Theorem, the root must lie in the interval $(\frac{1}{4},\frac{1}{2})$.

Finally, we calculate the abscissa of point $P$. Since $P$ lies on the circle at a distance of $R=1$ from the center $C$ along the normal line, we use the parametric form $x_P=x_C+R\cos \theta$.

With $\cos \theta =\frac{-2t}{\sqrt{1+4t^2}}$, we find:

Substituting our approximate root $t\approx 0.31$, we get $x_P\approx 0.47$. This value sits perfectly within the interval $(\frac{1}{4},\frac{1}{2})$.