The shortest distance between the point (3/2, 0) and the curve y = √x, (x 0) is : (1) √5/2 (2) 5/4

Q: Wait, why did we use $t$ instead of $x$ for the point $Q$?

Great question! If we used $x$, our point would be $(x, \sqrt{x})$. When we plug this into the distance formula, we get a square root inside a square root, which is a nightmare to differentiate. By setting $y = t$, we get $x = t^2$. This transforms the curve into a polynomial form $(t^2, t)$, making the algebra clean and elegant.

Q: Why can we minimize $d^2$ instead of $d$?

Because the distance $d$ is always non-negative. If $d^2$ is minimized, then $d$ (which is $\sqrt{d^2}$) is also minimized at the exact same point. It is a standard, powerful trick in optimization to avoid dealing with the derivative of a square root function.

Q: Why complete the square instead of using differentiation?

Differentiation is the 'brute force' method that always works, but completing the square is the 'ninja' method. It reveals the structure of the function. It shows us exactly when the minimum occurs (when the squared term is zero) without the extra work of finding critical points and checking the second derivative.

Question

The shortest distance between the point $(\frac{3}{2}, 0)$ and the curve $y = \sqrt{x}, (x > 0)$ is :

Options:

(A) $\frac{\sqrt{5}}{2}$

(B) $\frac{5}{4}$

(C) $\frac{3}{2}$

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

Sigma Percentile · Accepted Answer

### The Geometry of Optimization: A Journey to the Shortest Path

Imagine you are standing on the Cartesian plane. You are anchored at the point $P(\frac{3}{2}, 0)$, and before you lies the graceful, sweeping arc of the curve $y = \sqrt{x}$.

Your mission is to find the shortest possible distance between your position and any point on that curve. This is a fundamental challenge of optimization that appears everywhere from physics to machine learning.

### Phase 1: The Art of Parameterization

Most students immediately try to write the distance formula using $x$ and $y$. They write $Q(x, \sqrt{x})$ and then try to minimize the distance formula. While this is mathematically correct, it is a trap.

The square root of $x$ makes the algebra messy and prone to errors. Instead, let us be clever and parameterize the curve. If we define the $y$-coordinate as a parameter $t$, then the $x$-coordinate must be $t^2$ (since $y = \sqrt{x}$ implies $x = y^2$).

Now, any point $Q$ on the curve can be represented as $Q(t^2, t)$. This simple shift turns a radical expression into a polynomial, which is much friendlier to manipulate.

### Phase 2: The Distance Formula

Now that we have our point $Q(t^2, t)$ and our fixed point $P(\frac{3}{2}, 0)$, we apply the distance formula:

$$d = \sqrt{(t^2 - \frac{3}{2})^2 + (t - 0)^2}$$

As we noted, working with the square root is unnecessary. Let us focus on the square of the distance, $d^2$, because the value of $t$ that minimizes $d^2$ will also minimize $d$.

So, we define our function as:

$$d^2 = (t^2 - \frac{3}{2})^2 + t^2$$

### Phase 3: The Algebraic Dance

Now, we expand the expression. Using the identity $(a - b)^2 = a^2 - 2ab + b^2$, we expand $(t^2 - \frac{3}{2})^2$ to get:

$$t^4 - 3t^2 + \frac{9}{4}$$

Adding the remaining $t^2$ term, our expression becomes:

$$d^2 = t^4 - 3t^2 + \frac{9}{4} + t^2$$

Combining the like terms, we arrive at the simplified quadratic-like form:

$$d^2 = t^4 - 2t^2 + \frac{9}{4}$$

### Phase 4: The Masterstroke

We could differentiate this with respect to $t$ and set it to zero, but there is a more elegant path. Look at the expression $t^4 - 2t^2$. It looks remarkably like the expansion of $(t^2 - 1)^2$, which is $t^4 - 2t^2 + 1$.

We can force this structure by splitting the constant $\frac{9}{4}$ into $1 + \frac{5}{4}$. Thus, our expression becomes:

$$d^2 = (t^4 - 2t^2 + 1) + \frac{5}{4}$$

This simplifies to:

$$d^2 = (t^2 - 1)^2 + \frac{5}{4}$$

This is the moment of clarity. Since $(t^2 - 1)^2$ is a square, its minimum value is $0$, which occurs when $t^2 = 1$. Therefore, the minimum value of $d^2$ is simply $\frac{5}{4}$.

### Conclusion

Finally, we return to the distance $d$. Since $d^2 = \frac{5}{4}$, we take the square root to find:

$$d = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$

We have successfully navigated the geometry, parameterized the curve, and used algebraic completion to find the shortest distance. The **final answer is $\frac{\sqrt{5}}{2}$**, which corresponds to option (A).

Remember, the most difficult problems often yield to the most elegant simplifications. Keep practicing, and keep looking for the structure hidden within the equations.

Animated Solution for Mathematics - Differentiation: The shortest distance between the point $(\frac{3}{2}, 0)$ and the curve $y = x, (x > 0)$ is :

Select Answer:

Visualized Solution

Visualizing the Curve and Point $P$

Parameterizing Point $Q$ on the Curve

Applying the Distance Formula

Substituting Coordinates into the Formula

Expanding the Squared Term

Simplifying the Distance Squared Expression

Completing the Square Strategy

Rewriting as a Perfect Square

Finding the Minimum Condition

Calculating the Shortest Distance

Final Answer and Summary

The Sigma Insight: Maxima and Minima

The Geometry of Optimization

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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