# 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 $P$.

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) Step 1: Let P(x_1,y_1) be a point on y=x^3 \large\frac{dy}{dx}$$=3x^2$
Slope of the tangent at $(x_1,y_1)=3{x_1}^2$
Step 2:
Equation of the tangent at $(x_1,y_1)$
$(y-y_1)=3{x_1}^2(x-x_1)$
$y-3{x_1}^2x-y_1+3{x_1}^3=0$
$y-3{x_1}^2x-{x_1}^3+3{x_1}^3=0$ (Since $(x_1y_1)$ does on the curve)
$y-3{x_1}^2x+2x_1^3=0$
Step 3:
At the P and Q $(x_2,y_2)$ Where the tangent intersects the curve again , $y_2={x_2} ^3$
Substituting $(x_2,{x_2}^3)$ in the equation of the tangent
${x_2}^3-3{x_1}^2x_2+2{x_1}^3=0$
${x_2}^3-3{x_1}^2x_2+3{x_1}^3-{x_1}^3=0$
$({x_2}^3-{x_1}^3)-3{x_1}^2(x_2-x_1)=0$
$(x_2-x_1)[{x_2}^2+x_1x_2+{x_1}^2-3{x_1}^2]=0$
Since $x_2 \neq x_1, {x_2}^2+x_1x_2-2x_1^2=0$
$(x_2-x_1)(x_2+2x_1)=0$
Since $x_2 \neq x_1$
$x_2=-2x_1$
Step 4:
Now the slope of the tangent at $(x_2,y_2)$
is $+3{x_2}^2=3(-2x_1)^2=12 {x_1}^2 =4 (3{x_1}^2)$
Therefore slope of the tangent at $(x_2,y_2)=4$ times
the slope of the tangent at $(x_1,y_1)$