Question

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 $P$.

Answer 1

Let $(a,a^3)$ be a point on the curve $y=x^3$ .....$\left(1\right)$
$\frac{dy}{dx}=3x^2\frac{dy}{dx}$ at $(a,a^3)=3a^2=m=$ Slope of tangent
Equation of tangent at $P$ is
$y-a^3=3a^2(x-a)$
(i.e.,) $y-a^3=3a^{2}x-3a^3$
$y=3a^{2}x-2a^3$ .....$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$x^3=3a^{2}x-2a^3\Rightarrow x^3-3a^{2}x-2a^3=0$
${(x-a)}^2(x+2a)=0$
$x=a$ or $2a$

At $x=a;y=a^3$ at $x=-2a,y=-8a^3$
But $(a,a^3)$ is the point $P$
$\therefore$ $Q=(-2a,-8a^3)$

$\frac{dy}{dx}$ at $Q$ $(-2a,-8a^3)=3(-2a^2)=12a^2$
$\frac{dy}{dx}$ at $P(a,a^3)=3{(a}^2)=3a^2$
$\therefore$ the slope of $Q=4$ (slope at $P$).