From the point (-2, 3) two tangents are drawn to the parabola y² = 8x…

Step 1. The slopes. $3 = -2m + \dfrac{2}{m}$, so $2m^2 + 3m - 2 = 0$. Then $(2m - 1)(m + 2) = 0$ and $m = \dfrac{1}{2}$ or $m = -2$. The product is $-1$, so the tangents are perpendicular, as they must be, since $(-2, 3)$ is on the directrix $x = -2$.

Step 2. The points of contact. For $m = \dfrac{1}{2}$: $\left(\dfrac{2}{1/4}, \dfrac{4}{1/2}\right) = (8, 8)$. For $m = -2$: $\left(\dfrac{2}{4}, \dfrac{4}{-2}\right) = (0.5, -2)$. Both are on $y^2 = 8x$.

Step 3. The chord. The chord of contact is $3y = 4(x - 2)$, which is $4x - 3y - 8 = 0$. It passes through both points. The distance between them is $\sqrt{7.5^2 + 10^2} = \sqrt{156.25} = 12.5$.