Let C1 and C2 be respectively, the parabolas x² = y - 1 and y² = x - 1. Let P be any point on C1 and

Q: Why is the shortest distance between two symmetric curves always perpendicular to the line of symmetry?

Because the line of symmetry acts as a mirror. Any path that is not perpendicular would be 'tilted' relative to the mirror, and by reflecting that path, you would find a shorter path that is perpendicular. It is the principle of reflection.

Q: Why can we just swap the coordinates of $P_0$ to get $Q_0$?

Because $C_2$ is the reflection of $C_1$ across $y = x$. If $P_0$ is on $C_1$, its reflection $P_0'$ must be on $C_2$. Since the distance is minimized at $P_0$, the corresponding point on $C_2$ must be its reflection.

Q: Can we use the distance formula directly?

You could, but it would involve a messy square root function of two variables. Using the symmetry and calculus is much more elegant and less prone to algebraic errors.

Question

Let $C_1$ and $C_2$ be respectively, the parabolas $x^2 = y - 1$ and $y^2 = x - 1$. Let $P$ be any point on $C_1$ and $Q$ be any point on $C_2$. Let $P_1$ and $Q_1$ be the reflections of $P$ and $Q$, respectively, with respect to the line $y = x$. Prove that $P_1$ lies on $C_2, Q_1$ lies on $C_1$ and $PQ \ge \min\{PP_1, QQ_1\}$. Hence or otherwise determine points $P_0$ and $Q_0$ on the parabolas $C_1$ and $C_2$ respectively such that $P_0Q_0 \le PQ$ for all pairs of points $(P, Q)$ with $P$ on $C_1$ and $Q$ on $C_2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, coordinate-plane landscape. Before you lie two elegant curves, two parabolas, $C_1$ and $C_2$.

The first, $C_1$, is defined by $x^2 = y - 1$, rising gracefully like a fountain. The second, $C_2$, is defined by $y^2 = x - 1$, sweeping out to the right like a breaking wave.

These curves are mirror images, dancing around the line $y = x$. This line is not just a boundary; it is the master key to our problem.

### The Proof of Reflection

Let us take a point $P(x_1, y_1)$ on $C_1$. Its existence is governed by the equation $x_1^2 = y_1 - 1$.

Now, reflect this point across the line $y = x$. The coordinates swap, giving us $P_1(y_1, x_1)$.

Does this point $P_1$ belong to $C_2$? If we substitute $y_1$ for $x$ and $x_1$ for $y$ in the equation for $C_2$, we get:

$$x_1^2 = y_1 - 1$$

This is identical to our original condition for $P$. Thus, the reflection of any point on $C_1$ is guaranteed to land on $C_2$.

### The Geometry of Distance

Consider the distance between any point $P$ on $C_1$ and any point $Q$ on $C_2$. We are told that $PQ \ge \min\{PP_1, QQ_1\}$.

Think of the line $y = x$ as a perpendicular bisector. The geometry of the trapezoids formed by these points and their reflections forces the distance $PQ$ to be constrained by the distance between the curves themselves.

### The Calculus of Minimization

Our ultimate goal is to find the points $P_0$ and $Q_0$ that minimize this distance. The shortest path between two symmetric curves must lie along their common normal.

Because of the symmetry across $y = x$, this common normal must be perpendicular to the line $y = x$. If the normal is perpendicular to a line with slope $1$, the normal itself must have a slope of $-1$.

Consequently, the tangent at our ideal point $P_0$ must be parallel to the line $y = x$. This means the slope of the tangent at $P_0$ must be exactly $1$.

### The Final Calculation

We have our condition: the slope of the tangent at $P_0$ is $1$. We take the equation of $C_1$, which is $y = x^2 + 1$, and differentiate it with respect to $x$.

The derivative, $\frac{dy}{dx}$, is $2x$. Setting this equal to $1$, we find:

$$2x = 1 \Rightarrow x = \frac{1}{2}$$

Substituting this back into the equation for $C_1$, we find:

$$y = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}$$

Thus, the point $P_0$ is:

$$P_0 = \left(\frac{1}{2}, \frac{5}{4}\right)$$

By the symmetry we proved earlier, $Q_0$ is simply the reflection of $P_0$. Therefore, the coordinates of $Q_0$ are:

$$Q_0 = \left(\frac{5}{4}, \frac{1}{2}\right)$$

We have arrived at our destination through the elegant application of symmetry and calculus. You have mastered the geometry of the parabolas.

Visualized Solution

The Geometric Setup

Proving Symmetry

The Distance Inequality

Minimizing the Distance

Setting up the Slope

Differentiating $C_{1}$

Solving for the $x$ -coordinate

Finding the $y$ -coordinate

Finding $Q_{0}$ and Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Proof of Reflection

The Geometry of Distance

The Calculus of Minimization

The Final Calculation

Similar Questions

Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

Let the image of parabola $x^{2} = 4 y$ , in the line $x - y = 1$ be $(y + a)^{2} = b (x - c)$ , $a, b, c \in N$ . Then $a + b + c$ is equal to

Let the curve $C$ be the mirror image of the parabola $y^{2} = 4 x$ with respect to the line $x + y + 4 = 0$ . If $A$ and $B$ are the points of intersection of $C$ with the line $y = - 5$ , then the distance between $A$ and $B$ is

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

For any real $t$ , $x = \frac{e ^{t} + e ^{- t}}{2}$ , $y = \frac{e ^{t} - e ^{- t}}{2}$ is a point on the hyperbola $x^{2} - y^{2} = 1$ . Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to $t_{1}$ and $- t_{1}$ is $t_{1}$ .

Let $O$ be the vertex of the parabola $x^{2} = 4 y$ and $Q$ be any point on it. Let the locus of the point $P$ , which divides the line segment $O Q$ internally in the ratio $2 : 3$ be the conic $C$ . Then the equation of the chord of $C$ , which is bisected at the point $(1, 2)$ , is :

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then