Q.Let P be a point on the curve $y = x^3$ and suppose that the tangent line at P intersects the curve again at Q. Prove that the slope at Q is four times the slope at P.

Finding the tangent at a general point $P=(p,p^3)$ on $y=x^3$ and its second intersection $Q$ with the curve shows the slope at $Q$ is exactly $4$ times the slope at $P$.

1. Let $P=(p,p^3)$ be a point on $y=x^3$.
2. Differentiating, $\dfrac{dy}{dx}=3x^2$, so the slope of the tangent at $P$ is $m_P = 3p^2$.
3. Equation of tangent at $P$: $y-p^3 = 3p^2(x-p) \Rightarrow y = 3p^2x - 3p^3+p^3 = 3p^2x-2p^3$.
4. Find where this line meets the curve $y=x^3$ again: set $x^3 = 3p^2x - 2p^3$, i.e. $x^3-3p^2x+2p^3=0$.
5. Since the tangent touches the curve at $x=p$, $x=p$ is a (double) root of this cubic. Factor it out: dividing $x^3-3p^2x+2p^3$ by $(x-p)^2=x^2-2px+p^2$ gives quotient $(x+2p)$. Check: $(x-p)^2(x+2p) = (x^2-2px+p^2)(x+2p) = x^3+2px^2-2px^2-4p^2x+p^2x+2p^3 = x^3-3p^2x+2p^3$ ✓. …