A straight line graph, obtained by plotting $\frac{x}{y}$ against $x^2$, passes through the points $(-2, 5)$ and $(4, 8)$. Prove that $y = \frac{2x}{x^2 + 12}$.

1. Calculate the gradient of the straight line

Let $Y = \frac{x}{y}$ and $X = x^2$.
The line passes through $(X_1, Y_1) = (-2, 5)$ and $(X_2, Y_2) = (4, 8)$.
Gradient $m = \frac{8 - 5}{4 - (-2)} = \frac{3}{6} = \frac{1}{2}$

2. Find the equation of the straight line in terms of $X$ and $Y$

$Y - 5 = \frac{1}{2}(X - (-2))$
$Y = \frac{1}{2}X + 1 + 5$
$Y = \frac{1}{2}X + 6$

3. Substitute the original variables back into the equation

Substitute $Y = \frac{x}{y}$ and $X = x^2$:
$\frac{x}{y} = \frac{1}{2}x^2 + 6$

4. Rearrange the equation to make $y$ the subject

$\frac{x}{y} = \frac{x^2 + 12}{2}$
$y = \frac{2x}{x^2 + 12}$ (Shown)