Find the shortest distance of the point (0, c) from the parabola y = x² where 0 ≤ c ≤ 5.

Q: Wait, why do we minimize the squared distance $D^2$ instead of the distance $D$ itself?

That is a fantastic question! Minimizing $D$ involves dealing with square roots, which makes the differentiation process messy and prone to algebraic errors. Since the distance $D$ is always non-negative, minimizing $D^2$ yields the exact same $y$-coordinate for the minimum as minimizing $D$. It is a standard, powerful technique in optimization to simplify the function before differentiating.

Q: What happens if $c < 1/2$? Why does the formula change?

Great observation! If $c < 1/2$, the critical point we found, $y = c - 1/2$, becomes negative. However, for our parabola $y = x^2$, $y$ must be greater than or equal to zero. When $c$ is small, the 'normal' condition doesn't occur on the parabola itself; instead, the closest point is simply the vertex at the origin $(0,0)$. In that case, the distance is just the distance from $(0, c)$ to $(0, 0)$, which is $c$.

Q: Why is the normal condition so important in this problem?

The normal condition is the geometric soul of this problem. If you draw a circle centered at $P(0, c)$ with radius $r$, the circle will touch the parabola at the point of minimum distance. At this point of contact, the tangent to the circle (which is perpendicular to the radius) must coincide with the tangent to the parabola. Therefore, the radius (the line $PQ$) must be perpendicular to the tangent of the parabola, making it the normal line.

Question

Find the shortest distance of the point $(0, c)$ from the parabola $y = x^2$ where $0 \le c \le 5$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at a point $P(0, c)$ on the $y$-axis, looking down at the elegant curve of the parabola $y = x^2$. You want to find the shortest path to reach the curve.

We have a fixed point $P(0, c)$ and a moving point $Q(x, y)$ on the parabola. The distance $D$ between them is given by the distance formula:

$$D = \sqrt{(x - 0)^2 + (y - c)^2}$$

This is our starting point, but the square root is a heavy burden to carry through differentiation.

### The Power of Substitution

We know that for any point $Q$ on the parabola, $y = x^2$. This is the key that unlocks the problem. Instead of juggling two variables, $x$ and $y$, we can eliminate $x$ entirely.

Since $x^2 = y$, our distance formula becomes much cleaner. We choose to minimize the squared distance $D^2$ because the value of $y$ that minimizes $D^2$ will also minimize $D$.

We define our function:

$$D^2 = x^2 + (y - c)^2$$

Substituting $x^2 = y$, we get $D^2 = y + (y - c)^2$. Expanding this, we arrive at a beautiful quadratic function:

$$f(y) = y^2 + (1 - 2c)y + c^2$$

### The Calculus Engine

Now, we let calculus do the heavy lifting. To find the minimum of this quadratic function, we take the derivative with respect to $y$:

$$f'(y) = \frac{d}{dy}[y^2 + (1 - 2c)y + c^2]$$

This yields $f'(y) = 2y + (1 - 2c)$. Setting this derivative to zero, we find the critical point:

$$2y + 1 - 2c = 0 \Rightarrow y = c - \frac{1}{2}$$

This $y$-coordinate represents the vertical position on the parabola that is closest to our point $P$.

### The Final Reveal

With our optimal $y$ in hand, we substitute it back into our squared distance formula to find the minimum value:

$$D^2 = \left(c - \frac{1}{2}\right) + \left(c - \frac{1}{2} - c\right)^2$$

Watch the magic happen as the $c$ terms cancel out:

$$D^2 = \left(c - \frac{1}{2}\right) + \left(-\frac{1}{2}\right)^2$$

This simplifies to $D^2 = c - \frac{1}{2} + \frac{1}{4}$, which results in:

$$D^2 = c - \frac{1}{4}$$

Finally, taking the square root, we find the shortest distance:

$$D = \sqrt{c - \frac{1}{4}}$$

This result is valid for **$c \ge \frac{1}{2}$**. If $c$ is smaller, the geometry shifts, and the vertex becomes the closest point.

Animated Solution for Mathematics - Differentiation: Find the shortest distance of the point $(0, c)$ from the parabola $y = x^{2}$ where $0 \leq c \leq 5$ .

Visualized Solution

Visualizing the Problem

Defining a General Point $Q$

Simplifying the Distance Expression

Formulating the Function $f (y)$

Differentiating the Function $f (y)$

Finding the Critical Point

Substituting Back to Find Minimum $D^{2}$

Calculating the Shortest Distance

Geometric Interpretation & Constraints

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Power of Substitution

The Calculus Engine

The Final Reveal

Similar Questions

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

A helicopter is flying along the curve given by $y - x^{3/2} = 7, (x \geq 0)$ . A soldier positioned at the point $(1/2, 7)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

If $p (x)$ be a polynomial of degree 3 satisfying $p (- 1) = 10, p (1) = - 6$ and $p (x)$ has maxima at $x = - 1$ and $p^{'} (x)$ has minima at $x = 1$ . Find the distance between the local maxima and local minima of the curve.

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $y = 12 - x^{2}$ such that the rectangle lies inside the parabola, is :-

Find the coordinates of the point on the curve $y = \frac{x}{1 + x ^{2}}$ where the tangent to the curve has the greatest slope.

Let $A$ be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region $A$ is________

A point $P$ is given on the circumference of a circle of radius $r$ . Chord $QR$ is parallel to the tangent at $P$ . Determine the maximum possible area of the triangle $P QR$ .

Let $x$ and $y$ be two real variables such that $x > 0$ and $x y = 1$ . Find the minimum value of $x + y$ .

The area (in sq. units) of the largest rectangle $A B C D$ whose vertices $A$ and $B$ lie on the $x$ -axis and vertices $C$ and $D$ lie on the parabola, $y = x^{2} - 1$ below the $x$ -axis, is :

Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} ln x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the shortest distance of the point (0,c) from the parabola y=x2 where 0≤c≤5.

Visualized Solution

Visualizing the Problem

Defining a General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Simplifying the Distance Expression

Formulating the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(y)

Differentiating the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(y)