The problem gives a quadratic function $y = \frac{1}{4}x^2$ which is shifted. The intersection points of the shifted parabola and the x-axis are $(-2a, 0)$ and $(4a, 0)$, where $a$ is a positive constant. We need to find the equation of the parabola $y = f(x)$, the range of $x$ where $y \le 10a^2$, and finally, the value of $a$ when the length of the line segment cut off by $y = f(x)$ and $y = 10a$ is $10$.

1. Problem Description

The problem gives a quadratic function $y = \frac{1}{4}x^2$ which is shifted. The intersection points of the shifted parabola and the x-axis are $(-2a, 0)$ and $(4a, 0)$, where $a$ is a positive constant. We need to find the equation of the parabola $y = f(x)$, the range of $x$ where $y \le 10a^2$, and finally, the value of $a$ when the length of the line segment cut off by $y = f(x)$ and $y = 10a$ is $10$.

2. Solution Steps

(1) Since the x-intercepts of the parabola $y = f(x)$ are $-2a$ and $4a$, we can write $f(x)$ in the form:

$f(x) = \frac{1}{4}(x - 4a)(x + 2a)$

Comparing this to the given form $f(x) = \frac{A}{B}(x - Ca)(x + Da)$, we have $A = 1, B = 4, C = 4, D = 2$.

(2) We want to find the range of $x$ such that $f(x) \le 10a^2$.

$\frac{1}{4}(x - 4a)(x + 2a) \le 10a^2$

$(x - 4a)(x + 2a) \le 40a^2$

$x^2 - 2ax - 8a^2 \le 40a^2$

$x^2 - 2ax - 48a^2 \le 0$

Comparing this to the given form $x^2 - Eax - FGa^2 \le 0$, we have $E = 2, FG = 48$.

Now we solve the inequality $x^2 - 2ax - 48a^2 \le 0$.

The roots of the quadratic equation $x^2 - 2ax - 48a^2 = 0$ are:

$x = \frac{2a \pm \sqrt{(-2a)^2 - 4(1)(-48a^2)}}{2} = \frac{2a \pm \sqrt{4a^2 + 192a^2}}{2} = \frac{2a \pm \sqrt{196a^2}}{2} = \frac{2a \pm 14a}{2}$

So, $x = \frac{2a + 14a}{2} = 8a$ or $x = \frac{2a - 14a}{2} = -6a$.

Since the parabola opens upwards, the inequality $x^2 - 2ax - 48a^2 \le 0$ is satisfied for $-6a \le x \le 8a$.

Comparing this with $-Ha \le x \le Ia$, we have $H = 6, I = 8$.

(3) The length of the line segment cut off by $y = f(x)$ and $y = 10a$ is $10$. This means that the difference between the $x$ values when $f(x) = 10a$ is $10$.

$\frac{1}{4}(x - 4a)(x + 2a) = 10a$

$x^2 - 2ax - 8a^2 = 40a$

$x^2 - 2ax - 8a^2 - 40a = 0$

$x = \frac{2a \pm \sqrt{4a^2 - 4(-8a^2 - 40a)}}{2} = \frac{2a \pm \sqrt{4a^2 + 32a^2 + 160a}}{2} = \frac{2a \pm \sqrt{36a^2 + 160a}}{2} = a \pm \sqrt{9a^2 + 40a}$

$x_1 = a + \sqrt{9a^2 + 40a}, x_2 = a - \sqrt{9a^2 + 40a}$.

Then, $|x_1 - x_2| = 10$.

$|a + \sqrt{9a^2 + 40a} - (a - \sqrt{9a^2 + 40a})| = 10$

$2\sqrt{9a^2 + 40a} = 10$

$\sqrt{9a^2 + 40a} = 5$

$9a^2 + 40a = 25$

$9a^2 + 40a - 25 = 0$

Therefore $2\sqrt{9a^2 + 40a} = 10$ matches $J\sqrt{Ka^2+LMa}=10$. So $J=2$, $K=9$, $LM=40$.

Now, solve for a:

$(9a - 5)(a + 5) = 0$

Since $a$ is positive, $a = \frac{5}{9}$.

Comparing $a = \frac{N}{O}$, we have $N = 5, O = 9$.

3. Final Answer

A = 1

B = 4

C = 4

D = 2

E = 2

FG = 48

H = 6

I = 8

J = 2

K = 9

LM = 40

N = 5

O = 9