The shortest distance between the line y = x and the curve y² = x - 2 is : (A) 7/(4√2) (B) 7/8 (C) 1

Q: Wait, why does the shortest distance have to be along the normal? Can't it just be any perpendicular line?

That is a fantastic question! Think of it this way: if you draw a line from the curve to the target line that is *not* normal, you can always 'slide' your starting point along the curve to make the segment shorter. The only point where you cannot 'slide' to reduce the distance further is when the segment is perfectly perpendicular to the tangent—which is the normal!

Q: Why do we differentiate $y^2 = x - 2$ as $2y \frac{dy}{dx} = 1$? I always get confused with the chain rule.

It is completely normal to feel that way! Remember, $y$ is a function of $x$. When you differentiate $y^2$ with respect to $x$, you are essentially treating $y$ as an inner function. So, the derivative of $y^2$ is $2y$ (the outer derivative) multiplied by $\frac{dy}{dx}$ (the derivative of the inner function $y$). Always keep that $\frac{dy}{dx}$ term attached!

Q: Why is the denominator in the distance formula $\sqrt{2}$? I thought it was just $\sqrt{A^2 + B^2}$.

You are absolutely correct! The formula is indeed $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. In our case, the line is $x - y = 0$, so $A = 1$ and $B = -1$. Plugging those in, we get $\sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$. It is just a simple application of the formula!

Question

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

Options:

(A) $\frac{7}{4\sqrt{2}}$

(B) $\frac{7}{8}$

(C) $\frac{11}{4\sqrt{2}}$

(D) $2$

Sigma Percentile · Accepted Answer

### The Dance of Geometry: Finding the Shortest Path

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey of geometric intuition. We have a line, $y = x$, and a parabola, $y^2 = x - 2$.

At first glance, they seem like two separate entities, drifting in the Cartesian plane. But there is a hidden connection between them—a bridge of minimum distance that we are going to uncover.

### Phase 1: The Intuitive Leap

Imagine you are standing on the line $y = x$. You look across the plane at the parabola $y^2 = x - 2$. You want to reach the parabola, but you want to take the shortest possible path.

If you walk towards the parabola at an angle, you are wasting energy. The shortest path is always the one that hits the curve 'head-on.'

In the language of geometry, this means the path must be perpendicular to the tangent of the curve at the point of contact. This is the **'Common Normal' principle**. If we draw a tangent to the parabola at the point closest to our line, that tangent must be perfectly parallel to $y = x$. This is our key to the kingdom.

### Phase 2: The Calculus Bridge

Now, let us bring in the heavy artillery: Calculus. We know the slope of our line $y = x$ is $m = 1$.

Since our tangent must be parallel to this line, the slope of the tangent at our mystery point must also be $1$. How do we find the slope of the tangent to the parabola? We differentiate!

Taking the equation $y^2 = x - 2$, we differentiate both sides with respect to $x$. Remember the chain rule? It is the heartbeat of calculus. The derivative of $y^2$ is $2y \frac{dy}{dx}$, and the derivative of $x - 2$ is simply $1$. So, we have:

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

Rearranging this, we find the general slope expression for any point on the parabola:

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

This expression is powerful. It tells us the slope of the tangent at any point $(x, y)$ on the curve. We set this equal to $1$ because we need our tangent to be parallel to $y = x$:

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

Just like that, we have found the $y$-coordinate of our point of contact!

### Phase 3: The Final Calculation

With $y = \frac{1}{2}$ in hand, finding the $x$-coordinate is a simple matter of substitution. Plugging $y = \frac{1}{2}$ back into our parabola equation $y^2 = x - 2$, we get:

$$\left(\frac{1}{2}\right)^2 = x - 2 \implies \frac{1}{4} = x - 2 \implies x = 2 + \frac{1}{4} = \frac{9}{4}$$

Our point of contact, let's call it $P$, is $\left(\frac{9}{4}, \frac{1}{2}\right)$. Now, we are at the finish line. We need the perpendicular distance from this point $P$ to the line $x - y = 0$.

We use the classic distance formula:

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

Substituting $A = 1$, $B = -1$, $C = 0$, and our point $\left(\frac{9}{4}, \frac{1}{2}\right)$:

$$d = \frac{|(1)(\frac{9}{4}) + (-1)(\frac{1}{2})|}{\sqrt{1^2 + (-1)^2}} = \frac{|\frac{9}{4} - \frac{2}{4}|}{\sqrt{2}} = \frac{\frac{7}{4}}{\sqrt{2}}$$

And there it is: the shortest distance is **$\frac{7}{4\sqrt{2}}$**.

### The Takeaway

Look at what we just did. We didn't just crunch numbers; we used the properties of tangents, the power of the chain rule, and the elegance of the distance formula to solve a spatial problem.

Whenever you face a 'shortest distance' problem in JEE, don't panic. Visualize the normal, find the slope, and let the calculus guide you home. You have the tools; now go out and conquer the next one!

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

Select Answer:

Visualized Solution

Visualize the Geometry

The Shortest Distance Principle

Slope of the Given Line

Differentiating the Curve

Finding the Slope Expression

Solving for the y-coordinate

Finding the x-coordinate

Identifying Point P

The Distance Formula

Substitution in Formula

Simplifying the Numerator

Final Calculation

Conclusion \& Takeaway

The Sigma Insight: Equation of Tangent and Normal

The Dance of Geometry

Phase 1

Phase 2

Phase 3

The Takeaway

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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 the line y=x and the curve y2=x−2 is :