<Question>
The smallest distance between the origin and a point on the graph of $y=\frac{1}{2}x^2-9$ can be expressed as $a$.  Find $a^2$.
</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 - 9, and then express that distance as 'a' and find a². Hmm, let's start by recalling how to calculate the distance between two points. The distance formula between the origin (0,0) and a point (x, y) is sqrt[(x - 0)² + (y - 0)²], which simplifies to sqrt(x² + y²). Since we need the smallest distance, we can work with the square of the distance to make things simpler, right? Because the square root function is monotonically increasing, the minimum of sqrt(x² + y²) occurs at the same x-value as the minimum of x² + y². So, maybe I can minimize x² + y² instead of the distance itself. That should save some steps.

Given that y = (1/2)x² - 9, I can substitute that into the distance squared formula.
So, let's denote D² = x² + y². Substituting y, we get D² = x² + [(1/2)x² - 9]².

Now, I need to find the minimum value of this function. Let's call this function f(x) = x² + [(1/2)x² - 9]². To find the minimum, I should take the derivative of f(x) with respect to x, set it equal to zero, and solve for x. Then verify it's a minimum using the second derivative or some other method.

Let me compute the derivative step by step.

First, expand the [(1/2)x² - 9]² term.
That would be (1/2 x² - 9)² = (1/2 x²)^2 - 2*(1/2 x²)*(9) + 9² = (1/4)x⁴ - 9x² + 81. So, f(x) becomes x² + (1/4)x⁴ - 9x² + 81. Combining like terms: x² - 9x² is -8x², so f(x) = (1/4)x⁴ - 8x² + 81.

Now, take the derivative f’(x).

<prompt_1.1> Compute the derivative of the function f(x) = (1/4)x⁴ - 8x² + 81. </prompt_1.1>
-----

Wait, but let me think again. The problem says the smallest distance can be expressed as 'a', find a². So, if the minimal distance squared is 40, then a² is 40. Therefore, the answer is 40. But let me just make sure there's no mistake in the derivative calculation.

Alternatively, maybe parametrize the distance squared and use another method, like completing the square or something else. Let me see.

<prompt_1.2> Consider alternative methods to verify the result of minimizing the distance squared. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The derivative of (1/4)x⁴ is x³. The derivative of -8x² is -16x. The derivative of 81 is 0. So, f’(x) = x³ - 16x.
</execution_1.1>

<execution_1.2>
Alternatively, since the function is in terms of x², maybe substitute t = x², then f(t) = t + [(1/2)t -9]^2. Let's compute that.

Let t = x², t ≥ 0. Then f(t) = t + [(1/2)t -9]^2. Expand the square term: [(1/2)t -9]^2 = (1/4)t² - 9t +81. Therefore, f(t) = t + (1/4)t² -9t +81 = (1/4)t² -8t +81. Now, this is a quadratic in terms of t. To find the minimum, since the coefficient of t² is positive (1/4), the parabola opens upwards, so the vertex is the minimum. The vertex occurs at t = -b/(2a) where a=1/4, b=-8. So t = -(-8)/(2*(1/4)) = 8/(1/2) = 16. So t=16. Then substitute back into f(t): (1/4)(16)^2 -8*(16) +81. Compute each term: (1/4)(256) = 64, 64 -128 +81 = (64 -128) +81 = (-64) +81 = 17. Wait, that's conflicting with the previous result. Hmm, this suggests that the minimal value is 17. But that contradicts the earlier result of 40. So which one is correct?

Wait, hold on. There's a mistake here. Let me check.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Set this equal to zero: x³ - 16x = 0. Factor out an x: x(x² - 16) = 0. So, the critical points are x = 0 or x² - 16 = 0, which gives x = ±4.

Now, we need to determine which of these critical points gives the minimum value. Let's evaluate f(x) at each critical point.

<prompt_2.1> Evaluate the function f(x) at the critical points x = 0, x = 4, and x = -4. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
First, x = 0: f(0) = (1/4)(0)^4 - 8*(0)^2 + 81 = 0 - 0 + 81 = 81.

Next, x = 4: f(4) = (1/4)(4)^4 - 8*(4)^2 + 81. Compute each term:
(1/4)(256) = 64,
-8*(16) = -128,
So, 64 - 128 + 81 = (64 - 128) + 81 = (-64) + 81 = 17.

Similarly, x = -4: Since the function f(x) is even (all terms are even powers of x), f(-4) will be the same as f(4), which is 17.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but that contradicts my previous calculation where I thought x=4 gives f(x)=40. Hmm, so which one is correct? Let me check my steps again.

<prompt_3.1> Verify the calculations for f(4) and confirm the minimum value. </prompt_3.1>
-----

Based on execution_1.2:
Let's recalculate:

<prompt_3.2> Recalculate the minimum value using the substitution method and identify the error. </prompt_3.2>

But according to the previous calculation, the minimal distance squared is 17, so the minimal distance is sqrt(17), so a² would be 17.
But which one is correct? There's a discrepancy here. Let me check where I made the mistake.

<prompt_3.3> Identify and correct the error in the substitution method and confirm the correct result. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, when I first computed f(4), I thought I had 40, but now it's 17. Let me check again. Wait, original f(x) after substitution was x² + y². Let me compute that directly for x=4. y = (1/2)(16) -9 = 8 -9 = -1. So, the point is (4, -1). Then the distance squared is 4² + (-1)^2 = 16 + 1 = 17. So that's correct. Then why did I think earlier that I had 40? Maybe I confused the distance with the distance squared. So, in the first approach, when I computed f(x) as (1/4)x⁴ -8x² +81, and found f(4)=17, which matches the direct computation. So the minimal distance squared is 17, so a²=17. But wait, the problem says "the smallest distance between the origin and a point on the graph... can be expressed as a. Find a²". So, if the minimal distance is sqrt(17), then a² is 17.
</execution_3.1>

<execution_3.2>
f(4) = (1/4)(4)^4 -8*(4)^2 +81.

(4)^2 = 16, so (4)^4 = 256. Then (1/4)*256 = 64. Then 8*(4)^2 = 8*16=128. So 64 -128 +81 = (64 +81) -128 = 145 -128 =17.
</execution_3.2>

<execution_3.3>
Wait, when I substituted t = x², then f(t) = t + [(1/2)t -9]^2. Then expanding [(1/2)t -9]^2: (1/4)t² -9*(1/2)t*2 + 81? Wait, no. Wait, (a - b)^2 = a² - 2ab + b². So here, a = (1/2)t, b =9. So, (1/2 t)^2 - 2*(1/2 t)*9 + 9² = (1/4)t² -9t +81. That's correct. Then f(t) = t + (1/4)t² -9t +81 = (1/4)t² -8t +81. That's correct. Then vertex at t = -b/(2a) = 8/(2*(1/4)) = 8/(1/2) = 16. So t=16. Then f(16) = (1/4)(256) -8*(16) +81 = 64 -128 +81 = 17. So the minimal value is 17, which is f(t) = x² + y² =17. Therefore, the minimal distance squared is 17, so a²=17. Then why did the calculus approach give 17 as the minimal value? Wait, earlier when I did calculus, I had f(x) = (1/4)x⁴ -8x² +81. Then f’(x)=x³ -16x. Setting to zero gives x=0, x=4, x=-4. Then f(4)= (1/4)(256) -8*(16) +81 =64 -128 +81=17. So that's the same as the substitution method. So earlier, I must have miscalculated. Wait, initially, I thought f(4) was 40, but that was wrong. So the correct minimal value is 17. Therefore, the answer is 17. But why did I think it was 40 earlier?

Wait, maybe in the beginning, I miscalculated when expanding the terms. Let me re-examine the substitution step.

Original f(x) = x² + [(1/2)x² -9]^2. Let me compute that again. Let me denote z = (1/2)x² -9. Then f(x) = x² + z². If z = (1/2)x² -9, then z² = [(1/2)x² -9]^2. Let's compute z²:

[(1/2)x² -9]^2 = ( (x²)/2 -9 )² = (x²/2)^2 - 2*(x²/2)*9 + 9² = x⁴/4 -9x² +81. Therefore, f(x) = x² + x⁴/4 -9x² +81 = x⁴/4 -8x² +81. So that's correct. Then taking derivative: f’(x) = x³ -16x. Setting to zero gives x=0, x=±4. Then f(4)= (4)^4 /4 -8*(4)^2 +81 = (256)/4 -8*16 +81 =64 -128 +81=17. So, yes, 17 is correct.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
So, the answer is 17.

But let me just confirm once more with another approach. Let's parametrize the parabola y=(1/2)x² -9. The distance squared from the origin is x² + y² =x² + ( (1/2)x² -9 )². Let me set u =x², so u ≥0.

<prompt_4.1> Use a parametrization approach to confirm the minimum distance squared. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Then the distance squared becomes u + ( (1/2)u -9 )². Let's compute that:

u + ( (u/2 -9) )² = u + u²/4 -9u +81 = u²/4 -8u +81. Now, this is a quadratic in u, which opens upwards (since coefficient of u² is positive). The vertex is at u = -b/(2a) = 8/(2*(1/4)) =8/(1/2)=16. So u=16. Then plug back into the quadratic: (16)^2 /4 -8*(16) +81=256/4 -128 +81=64 -128 +81=17. Same result. So, all methods lead to 17.

Therefore, the minimal distance squared is 17, so a²=17.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But just to be thorough, let's check if the point at x=4 is indeed on the parabola. y=(1/2)(16)-9=8-9=-1. So the point is (4,-1). The distance from (0,0) to (4,-1) is sqrt(16 +1)=sqrt(17). So that's correct.

Is there a possibility that there's another critical point that we missed? For example, when we took the derivative of f(x)=x⁴/4 -8x² +81, the derivative is f’(x)=x³ -16x. Setting to zero gives x(x² -16)=0, so x=0 or x=±4. These are the only real roots. So, no other critical points. Therefore, the minimal value is indeed 17.

Therefore, the answer is 17.

**Final Answer**
The value of \(a^2\) is \boxed{17}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Thus, the smallest distance squared is $\boxed{17}$.
</Final_answer>
</think>