To find the nearest distance from the origin to the parabola defined by the equation:

$y = (5 - x)^2 + 1$

we need to minimize the distance between a point on the parabola and the origin $(0, 0)$. The distance $d$ from a point $(x, y)$ on the parabola to the origin is given by the distance formula:

$d(x, y) = \sqrt{x^2 + y^2}$

Substituting the given equation of the parabola for $y$:

$y = (5 - x)^2 + 1$

The distance function becomes:

$d(x) = \sqrt{x^2 + \left( (5 - x)^2 + 1 \right)^2}$

Now, we minimize this function with respect to $x$ to find the point where the distance is shortest. I'll proceed with calculating the derivative of $d(x)$, set it to zero, and solve for $x$. Let me compute this.The real critical point that minimizes the distance is $x = 4$. The other two solutions involve complex numbers, which are not relevant in this context.

Now, I'll compute the corresponding $y$-value for $x = 4$ and then find the shortest distance from the origin to the point $(4, y)$ on the parabola.At $x = 4$, the corresponding $y$-value is 2, making the point on the parabola $(4, 2)$. The shortest distance from the origin to this point is:

$\text{Distance} = 2\sqrt{5} \approx 4.47$

Would you like more details or clarifications on any step?

Here are 5 related questions:

1. How do you find the vertex of a parabola?
2. What is the geometric significance of the derivative when minimizing distance?
3. Can this method be used for minimizing distances to any curve?
4. How do you calculate distances in higher-dimensional spaces?
5. What happens if we want the maximum distance instead of the minimum?

Tip: Minimizing or maximizing a function often involves setting the derivative equal to zero and analyzing the critical points.