A designer in Seoul draws the parabola y = x², with lengths in centimetres…

Step 1. The tangent at $t$. The gradient at $(t, t^2)$ is $2t$, so the line is $y - t^2 = 2t(x - t)$, that is $y = 2tx - t^2$.

Step 2. Pass through $P$. $-3 = 2t - t^2$, so $t^2 - 2t - 3 = 0$, which factorises as $(t - 3)(t + 1) = 0$. So $t = 3$ or $t = -1$.

Step 3. The gradients. $2t = 6$ and $2t = -2$.

Step 4. The points of contact. $A(3, 9)$ and $B(-1, 1)$. Their distance is $\sqrt{4^2 + 8^2} = \sqrt{80} = 8.944\ \text{cm}$.

Step 5. The area. The area formula gives $1(9 - 1) + 3(1 + 3) + (-1)(-3 - 9) = 32$. The area is half of this: $16\ \text{cm}^2$.

Why no tangent from $(0, 1)$. The equation $1 = -t^2$ has no real solution.