The shortest distance between the line y - x = 1 and the curve x = y² is (a) 2√3/8 (b) 3√2/5 (c) √3/

Q: Wait, why do we reject the negative root in the distance formula?

We don't actually reject a root; we use the absolute value. The distance formula $D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ involves a modulus. In our case, the expression inside is $a^2 - a + 1$. Since its discriminant $\Delta = (-1)^2 - 4(1)(1) = -3$ is negative, the quadratic is always positive for all real $a$. Thus, the modulus is redundant, and we can simply write $D = \frac{a^2 - a + 1}{\sqrt{2}}$.

Q: Why can't we just use the distance from the vertex $(0,0)$ to the line?

That is a common trap! The vertex $(0,0)$ is just one point on the parabola. The shortest distance occurs at the point where the tangent to the parabola is parallel to the given line. The vertex is not necessarily that point. By using the parametric approach, we mathematically find the exact point that satisfies the parallel-tangent condition.

Q: Is there a faster way using calculus instead of completing the square?

Absolutely! You can define the distance function $f(a) = a^2 - a + 1$ and find its minimum by setting its derivative $f'(a) = 2a - 1$ to zero. This gives $a = 1/2$. Substituting this back into the distance formula gives the same result. Both methods are elegant; completing the square is often faster for simple quadratics, while calculus is a robust, universal tool.

Question

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

Options:

(A) $\frac{2\sqrt{3}}{8}$

(B) $\frac{3\sqrt{2}}{5}$

(C) $\frac{\sqrt{3}}{4}$

(D) $\frac{3\sqrt{2}}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a beautiful exploration of coordinate geometry. Today, we are not just solving a problem; we are uncovering the hidden relationship between a curve and a line.

We have a parabola, defined by $x = y^2$, opening gracefully to the right, and a straight line, $x - y + 1 = 0$. Our goal is to find the absolute shortest distance between them.

Imagine standing on the parabola, looking across at the line. Where is the closest point? This is the question that will guide our journey.

### The Power of Parametrization

To conquer this, we must be strategic. We could work with $x$ and $y$, but that is like trying to solve a puzzle with two hands tied behind our back.

Instead, let us use the power of parametrization. Since $x = y^2$, if we define our $y$-coordinate as a parameter $a$, then our $x$-coordinate is automatically $a^2$.

Thus, any point $P$ on our parabola can be represented as $P(a^2, a)$. This is a massive simplification! We have reduced a two-variable problem into a single-variable optimization problem.

### Bridging the Gap with the Distance Formula

Now, we need to connect our point $P(a^2, a)$ to the line $x - y + 1 = 0$. We reach into our mathematical toolkit and pull out the perpendicular distance formula:

$$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Here, our line is $1x - 1y + 1 = 0$, so $A=1$, $B=-1$, and $C=1$. Substituting our point $P(a^2, a)$ into this formula, we get:

$$D = \frac{|a^2 - a + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|a^2 - a + 1|}{\sqrt{2}}$$

### The Beauty of the Discriminant

Look closely at the expression inside the modulus: $a^2 - a + 1$. Is it possible for this to be negative?

Let us check the discriminant, $\Delta = b^2 - 4ac$. Here, $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3$.

Since the discriminant is negative and the leading coefficient is positive, this quadratic is strictly positive for all real values of $a$. We can safely remove the modulus sign, leaving our distance function as:

$$D(a) = \frac{a^2 - a + 1}{\sqrt{2}}$$

### The Final Minimization

To find the minimum distance, we must minimize the numerator $a^2 - a + 1$. We use the technique of completing the square:

$$a^2 - a + 1 = \left(a^2 - a + \frac{1}{4}ight) - \frac{1}{4} + 1 = \left(a - \frac{1}{2}ight)^2 + \frac{3}{4}$$

The minimum value of this expression occurs when the squared term is zero, which happens at $a = \frac{1}{2}$. At this point, the minimum value of the numerator is $\frac{3}{4}$.

Plugging this back into our distance equation, we get:

$$D_{min} = \frac{1}{\sqrt{2}} 	imes \frac{3}{4} = \frac{3}{4\sqrt{2}}$$

Finally, we rationalize the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$, yielding the final result:

**$$D_{min} = \frac{3\sqrt{2}}{8}$$**

### The Geometric Insight

Before we conclude, let us appreciate the elegance of what we just did. The shortest distance between a curve and a line always occurs along their common normal.

At the point $P(\frac{1}{4}, \frac{1}{2})$, the tangent to the parabola has a slope of $1$, which is exactly the slope of our line $y = x + 1$. They are perfectly parallel!

This is the geometric soul of the problem. Keep this visualization in your heart, and you will never fear such problems again.

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

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Point on the Parabola

Distance Formula Application

Substituting the Values

Analyzing the Quadratic Expression

Completing the Square

Finding the Minimum Distance

Final Calculation and Rationalization

Geometric Insight

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Power of Parametrization

Bridging the Gap with the Distance Formula

The Beauty of the Discriminant

The Final Minimization

The Geometric Insight

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