The parabola y² = 8x has the point (2, 4) on it…

Step 1. At $(2, 4)$. $4y = 4(x + 2)$, which is $y = x + 2$. The slope is $1$.

Step 2. Slope $2$. $y = 2x + \dfrac{2}{2} = 2x + 1$, so the $y$-intercept is $1$. The point of contact is $\left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right) = \left(\dfrac{2}{4}, 2\right) = (0.5, 2)$. Check: $2^2 = 8 \times 0.5$.

Step 3. Through $(-2, 0)$. $0 = -2m + \dfrac{2}{m}$, so $m^2 = 1$ and $m = \pm 1$. The larger slope is $1$.