The shortest distance between line y - x = 1 and curve x = y² is (A) 3√2/8 (B) 8/(3√2) (C) 4/√3 (D)

Q: Wait, why is the shortest distance always along the common normal?

Think of it this way: if you draw any other line segment from the curve to the line that is not the normal, you can always move along the curve to find a point that is slightly closer. The normal is the unique path where the distance function reaches its local minimum. It is the geometric manifestation of the derivative being zero.

Q: Why can't we just use the distance formula between the line and the parabola directly?

The distance formula $d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$ is for the distance from a *point* to a line. A parabola is a collection of infinite points. We must first identify the specific point on the parabola that is closest to the line, and *then* apply the distance formula to that point.

Q: Why do we differentiate with respect to $y$ instead of $x$?

You absolutely can differentiate with respect to $x$! If $x = y^2$, then $y = \sqrt{x}$ (for the upper half). Differentiating $y = x^{1/2}$ gives $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$. Since $y = \sqrt{x}$, this is $\frac{1}{2y}$. Both methods lead to the same result. Differentiating with respect to $y$ is just a bit cleaner algebraically here.

Question

The shortest distance between line $y - x = 1$ and curve $x = y^2$ is

Options:

(A) 3\sqrt{2}/8

(B) 8/(3\sqrt{2})

(C) 4/\sqrt{3}

(D) $\sqrt{3}/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, smooth landscape. To your right, a beautiful, sweeping parabola defined by $x = y^2$ curves gracefully. To your left, a straight, unyielding line $y - x = 1$ cuts across the horizon.

Your task is to find the absolute shortest path from the curve to the line. We are looking for the point on the parabola that "kisses" the line most closely.

### The Calculus Bridge

The key lies in the concept of the **common normal**. If you were to slide a line parallel to our given line $y - x = 1$ toward the parabola, the first point of contact would be the point of shortest distance.

At this exact point, the tangent to the parabola must be perfectly parallel to the line. Since the line $y - x = 1$ can be rewritten as $y = x + 1$, its slope is clearly $m = 1$.

Therefore, the slope of the tangent to our parabola at the point of closest approach must also be $1$.

### The Algebraic Hunt

Our curve is $x = y^2$. To find the slope of the tangent, we differentiate with respect to $y$:

$$\frac{dx}{dy} = 2y$$

This means the slope of the tangent, $\frac{dy}{dx}$, is the reciprocal:

$$\frac{dy}{dx} = \frac{1}{2y}$$

We know this slope must be $1$ to be parallel to our line. Setting the derivative equal to the slope:

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

With the $y$-coordinate in hand, finding the $x$-coordinate is straightforward. Substituting $y = \frac{1}{2}$ into $x = y^2$, we get:

$$x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

We have found our point of contact: $P\left(\frac{1}{4}, \frac{1}{2}\right)$.

### The Final Calculation

We have a point $P\left(\frac{1}{4}, \frac{1}{2}\right)$ and a line $x - y + 1 = 0$. The distance $d$ from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is given by the formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Plugging in our values, we get:

$$d = \frac{|\frac{1}{4} - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}}$$

Simplifying the numerator:

$$\frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4}$$

The denominator is $\sqrt{2}$. Thus, the distance is:

$$d = \frac{3/4}{\sqrt{2}} = \frac{3}{4\sqrt{2}}$$

Rationalizing the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$, we arrive at the final result:

**$$d = \frac{3\sqrt{2}}{8}$$**

Animated Solution for Mathematics - Conic Sections: The shortest distance between line $y - x = 1$ and curve $x = y^{2}$ is

Select Answer:

Visualized Solution

Visualizing the Problem

The Concept of Common Normal

Slope of the Given Line

Differentiating the Curve

Finding the $y$ -coordinate

Finding the $x$ -coordinate

The Distance Formula

Substituting the Point

Final Calculation

Rationalizing the Denominator

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus Bridge

The Algebraic Hunt

The Final Calculation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between line y−x=1 and curve x=y2 is