<Question>
The smallest distance between the origin and a point on the graph of $y=\frac{1}{\sqrt{2}}\left(x^2-3\right)$ can be expressed as $\sqrt{a}/b$, where $a$ and $b$ are positive integers such that $a$ is not divisible by the square of any integer greater than one.  Find $a+b$.
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest distance between the origin and a point on the graph of y = (1/√2)(x² - 3). The answer should be in the form √a / b, where a and b are positive integers, and a is square-free. Then I have to find a + b. Hmm, let me think step by step.

First, the distance between the origin (0,0) and a point (x, y) on the graph is given by the distance formula: √(x² + y²). Since we want the smallest distance, I need to minimize this distance. But since the square root function is increasing, minimizing the distance is equivalent to minimizing the square of the distance, which is x² + y². That might be easier to work with because it avoids dealing with the square root.

So, let's define the square of the distance as D = x² + y². Given that y = (1/√2)(x² - 3), I can substitute that into the equation for D. Let's do that:

<prompt_1.1> Define the square of the distance between the origin and a point on the graph using the given function. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
D = x² + [(1/√2)(x² - 3)]²

Let me compute that step by step. First, square the second term:

[(1/√2)(x² - 3)]² = (1/2)(x² - 3)²

So, expanding that, (x² - 3)² is x⁴ - 6x² + 9. Therefore, multiplying by 1/2 gives (1/2)x⁴ - 3x² + 9/2.

So now, D = x² + (1/2)x⁴ - 3x² + 9/2

Combine like terms:

x² - 3x² = -2x², so D = (1/2)x⁴ - 2x² + 9/2
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, to find the minimum of D, I need to take the derivative of D with respect to x, set it equal to zero, and solve for x. Then check if it's a minimum using the second derivative or some other method.

So, let's compute dD/dx:

<prompt_2.1> Compute the derivative of the squared distance function and set it to zero to find critical points. </prompt_2.1>
-----

Let me just check if there's another way to approach this problem, maybe parametrizing the distance and using calculus, but I think the method I used is standard for optimization problems involving distance from a point to a curve. The key steps are setting up the distance squared function, taking the derivative, and solving for critical points.

<prompt_2.2> Consider alternative methods and verify if the current solution is correct by analyzing the problem from different perspectives. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
dD/dx = (4x³)/2 - 2*2x /2? Wait, let me do it carefully.

Wait, D = (1/2)x⁴ - 2x² + 9/2

So derivative term by term:

- The derivative of (1/2)x⁴ is 4*(1/2)x³ = 2x³
- The derivative of -2x² is -4x
- The derivative of 9/2 is 0

So, dD/dx = 2x³ - 4x

Set this equal to zero for critical points:

2x³ - 4x = 0

Factor out 2x:

2x(x² - 2) = 0

So, the critical points are x = 0 or x² = 2, which gives x = √2 or x = -√2.
</execution_2.1>

<execution_2.2>
Also, checking the second derivative was important to confirm it's a minimum. Sometimes, if the second derivative is zero, you might need higher-order derivatives, but here it's straightforward. Since the function D is a quartic polynomial, which is convex, there should be only one minimum, so the critical point we found is indeed the global minimum.

Another way to check: maybe using geometry. The graph of y = (1/√2)(x² - 3) is a parabola opening upwards, vertex at (0, -3/√2). The origin is above the vertex. The closest point on the parabola to the origin should lie along the line perpendicular to the tangent of the parabola at that point. But that approach would involve more steps: finding the tangent line, then the perpendicular line, finding intersection with the parabola, etc. But since we already have the calculus method working, which is straightforward, maybe that's not necessary. However, just to satisfy my curiosity, let me try that approach quickly.

The slope of the tangent line to the parabola at point (x, y) is dy/dx = (1/√2)(2x) = √2 x. Therefore, the slope of the normal line (perpendicular) is -1/(√2 x). The equation of the normal line is then:

y - y1 = m(x - x1)

Where (x1, y1) is (x, (1/√2)(x² - 3)), and m is -1/(√2 x). So:

y - (1/√2)(x² - 3) = (-1/(√2 x))(x - x1)

Wait, x1 is x here. Let me write it properly:
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, we need to evaluate D at these critical points to find which one gives the minimum distance.

<prompt_3.1> Evaluate the squared distance at each critical point to determine the minimum distance. </prompt_3.1>

But let me verify this because sometimes endpoints or other critical points can be issues, but since the graph of the function y is a parabola extending to infinity, the minimal distance should occur at one of these critical points.
Wait, but let's confirm by checking the second derivative or the behavior of the first derivative.

Alternatively, since as x approaches infinity, D = (1/2)x⁴ - 2x² + 9/2 tends to infinity, and as x approaches zero, D = 0 + 0 + 9/2 = 9/2. So at x = 0, D is 9/2, which is 4.5. But at x = √2, D is 17/4 = 4.25, which is less than 4.5, so that's indeed a minimum. So that's correct.
-----

Based on execution_2.2:
Let's set y = (√2/2)(x² - 3) equal to this line. So:

(√2/2)(x² - 3) = (-1/(√2 x))(x - x) + [value at x? Wait, maybe I need to re-express this.

Wait, no. The normal line at (x, y) passes through the origin if the origin lies on the normal line. Wait, but we want the origin to be the closest point. Wait, perhaps the closest point is where the line connecting the origin to the parabola is perpendicular to the tangent line of the parabola at that point. So that's another way to phrase it, which is essentially what I started earlier.

So, the vector from the origin to the point (x, y) is (x, y), and the tangent vector to the parabola at that point is (1, dy/dx) = (1, √2 x). For these two vectors to be perpendicular, their dot product should be zero.

<prompt_3.2> Verify the closest point using the geometric approach by checking perpendicularity. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
First, x = 0:

D(0) = (1/2)(0)^4 - 2*(0)^2 + 9/2 = 9/2

Then, x = √2:

First, compute x⁴: (√2)^4 = (2)^(2) = 4

So, D(√2) = (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = (2 - 4) + 4.5 = (-2) + 4.5 = 2.5 = 5/2

Similarly, for x = -√2, since all terms are even powers or absolute values, it will be the same as x = √2, so D(-√2) is also 5/2.
</execution_3.1>

<execution_3.2>
So:

x * 1 + y * (√2 x) = 0

But y = (√2 / 2)(x² - 3). Substitute that in:

x + [(√2 / 2)(x² - 3)] * (√2 x) = 0

Let me compute that:

First, multiply √2 / 2 and √2 x:

(√2 * √2) / 2 * x = (2)/2 * x = x

So, the second term becomes x*(x² - 3)

Therefore, the equation is:

x + x(x² - 3) = 0

Factor out x:

x[1 + (x² - 3)] = 0

Simplify inside the brackets:

1 + x² - 3 = x² - 2

So, x(x² - 2) = 0

Which gives x = 0 or x² = 2, so x = √2 or x = -√2, same as before.

Therefore, the same critical points. Then, as before, x = 0 gives D = 9/2, and x = √2 gives D = 17/4. So, same result. Therefore, this method also leads to the same answer, which confirms that the minimal distance occurs at x = ±√2, and the minimal D is 17/4, so the distance is √(17/4) = √17 / 2. Therefore, a = 17 and b = 2, so a + b = 19.

This makes me more confident that the answer is indeed 19. Therefore, I think that's correct.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So, the minimal distance squared is 5/2, so the minimal distance is √(5/2) = √(10)/2. Wait, hold on: √(5/2) is equal to (√10)/2, right?

<prompt_4.1> Calculate the minimal distance from the squared distance and verify the calculations. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Because √(5/2) = √5 / √2 = (√5 * √2)/ (√2 * √2) = √10 / 2.

But the problem says the smallest distance can be expressed as √a / b, where a is not divisible by the square of any integer greater than one. Let's check a and b here.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
In √10 / 2, a is 10 and b is 2. Now, 10 is square-free since its prime factors are 2 and 5, neither of which is squared. So, a = 10, b = 2, so a + b = 12.

Therefore, the final answer is 12.

But let me double-check all steps to make sure I didn't make a mistake.

<prompt_5.1> Verify all calculations to ensure the solution is correct. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
First, the distance squared: D = x² + y². Substituted y = (1/√2)(x² - 3). Then squared that term:

[(1/√2)(x² - 3)]² = (1/2)(x² - 3)^2. Then expanded (x² - 3)^2 to x⁴ - 6x² + 9. Multiply by 1/2: (1/2)x⁴ - 3x² + 9/2. Then add x² to get D = (1/2)x⁴ - 2x² + 9/2. That seems correct.

Taking derivative: dD/dx = 2x³ - 4x. Setting to zero: 2x³ - 4x = 0 => 2x(x² - 2) = 0. Correct. So critical points at x = 0, x = √2, x = -√2. Then evaluated D at x = 0: 0 + 0 + 9/2 = 9/2. At x = √2: (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = 2.5 = 5/2. So D is 5/2 at those points. Then distance is √(5/2) = √10 / 2. Correct.

Then, confirming a and b: 10 and 2, which are coprime, and 10 is square-free. So a + b = 12. That seems right.

Alternatively, maybe check if there's another critical point or if I made a mistake in expanding (x² - 3)^2. Let's see: (x² - 3)^2 = x^4 - 6x² + 9. Yes. Multiply by 1/2: (1/2)x^4 - 3x² + 9/2. Correct. Then D = x² + that, so x² + (1/2)x^4 - 3x² + 9/2 = (1/2)x^4 - 2x² + 9/2. Correct.

Derivative: derivative of (1/2)x^4 is 2x³, derivative of -2x² is -4x, derivative of 9/2 is 0. So 2x³ - 4x. Correct. Then 2x(x² - 2) = 0. Correct. So x = 0 or x = ±√2. Correct.

Evaluating D at x = √2: x^4 is (√2)^4 = (2)^2 = 4. So (1/2)*4 = 2. Then -2x² is -2*(2) = -4. Then +9/2. So 2 - 4 + 4.5 = 2.5. Correct. So D = 5/2. So distance is √(5/2) = √10 / 2. Correct.

Therefore, the answer is 10 + 2 = 12.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the final answer is $\boxed{12}$.
</Final_answer>
</think>