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Problem :

Tangent at point $P_1$ other than $(0,0)$ on the curve $y =x^3$ meets the curve again at $P_2$ The tangent at $P_2$ meets the curve again at $P_3$ and so on. If $\frac{\text{area} (\Delta P_1P_2P_3)}{\text{area} (\Delta P_2P_3P_4)}=\lambda$ then find the value of $\lambda$

My approach :

If we want to find slope of tangent at point $(x_1,y_1)$ then $y =x^3$

$=\Rightarrow \frac{dy}{dx} =3x^2$

at point $(x_1,y_1)\ \frac{dy}{dx} =3x_1^2$

Please suggest how to get the point of intersection of the two curve and how to find the area as per the given problem . Will be of great help thanks.

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    $\begingroup$ Triangle in numerator is $\Delta P_2P_2P_3$? Are you sure? This does not exist. $\endgroup$ – SchrodingersCat Oct 20 '15 at 14:45
  • $\begingroup$ sorry that was P_1 $\endgroup$ – sultan Oct 20 '15 at 14:47
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Let $P_1(t,t^3)$ be a point on $y=x^3$ where $t\not=0$. Then, the equation of the tangent line of $y=x^3$ at $(t,t^3)$ is $y-t^3=3t^2(x-t)$. Then, we have $$x^3-t^3=3t^2(x-t)\iff (x-t)^2(x+2t)=0$$ So, it follows from this that we can have $$P_2(-2t,-8t^3),\quad P_3(4t,64t^3),\quad P_4(-8t,-512t^3).$$

Hence, $$\begin{align}\lambda&=\frac{[\triangle{P_1P_2P_3}]}{[\triangle{P_2P_3P_4}]}\\\\&=\frac{\frac 12|(-2t-t)(64t^3-t^3)-(-8t^3-t^3)(4t-t)|}{\frac 12|(4t+2t)(-512t^3+8t^3)-(64t^3+8t^3)(-8t+2t)|}\\\\&=\frac{|-3\times 63+27|}{|-6\times 504+72\times 6|}\\\\&=\color{red}{\frac{1}{16}}\end{align}$$

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