Application of Derivatives 1 Question 19
19. Tangent at a point $P _1$ {other than $\left.(0,0) }$ on the curve $y=x^{3}$ meets the curve again at $P _2$. The tangent at $P _2$ meets the curve at $P _3$ and so on.
Show that the abscissa of $P _1, P _2, P _3, \ldots, P _n$, form a GP. Also, find the ratio of
[area $\left.\left(\Delta P _1 P _2 P _3\right)\right] /\left[\operatorname{area}\left(\Delta P _2 P _3 P _4\right)\right]$.
(1993, 5M)
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Solution:
- Let any point $P _1$ on $y=x^{3}$ be $\left(h, h^{3}\right)$.
Then, tangent at $P _1$ is
It meets $y=x^{3}$ at $P _2$.
$$ y-h^{3}=3 h^{2}(x-h) $$
On putting the value of $y$ in Eq. (i), we get
$$ \begin{aligned} & x^{3}-h^{3}=3 h^{2}(x-h) \\ & \Rightarrow \quad(x-h)\left(x^{2}+x h+h^{2}\right)=3 h^{2}(x-h) \\ & \Rightarrow \quad x^{2}+x h+h^{2}=3 h^{2} \text { or } x=h \\ & \Rightarrow \quad x^{2}+x h-2 h^{2}=0 \\ & \Rightarrow \quad(x-h)(x+2 h)=0 \\ & \Rightarrow \quad x=h \text { or } x=-2 h \end{aligned} $$
Therefore, $x=-2 h$ is the point $P _2$,
which implies $y=-8 h^{3}$
Hence, point $P _2 \equiv\left(-2 h,-8 h^{3}\right)$
Again, tangent at $P _2$ is $y+8 h^{3}=3(-2 h)^{2}(x+2 h)$.
$$ \begin{array}{lrl} \text { It meets } & y & =x^{3} \text { at } P _3 \\ \Rightarrow \quad x^{3}+8 h^{3} & =12 h^{2}(x+2 h) \\ \Rightarrow \quad x^{2}-2 h x-8 h^{2} & =0 \\ \Rightarrow \quad(x+2 h)(x-4 h) & =0 \Rightarrow x=4 h \Rightarrow y=64 h^{3} \\ & \text { Therefore, } \quad P _3 & \equiv\left(4 h, 64 h^{3}\right) \\ & \text { Similarly, we get } \quad P _4 & \equiv\left(-8 h,-8^{3} h^{3}\right) \end{array} $$
Hence, the abscissae are $h,-2 h, 4 h,-8 h, \ldots$, which form a GP.
Let $D^{\prime}=\Delta P _1 P _2 P _3$ and $D^{\prime \prime}=\Delta P _2 P _3 P _4$
$$ \begin{aligned} \frac{D^{\prime}}{D^{\prime \prime}} & =\frac{\Delta P _1 P _2 P _3}{\Delta P _2 P _3 P _4}=\frac{\frac{1}{2}\left|\begin{array}{ccc} h & h^{3} & 1 \\ -2 h & -8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array}\right|}{\frac{1}{2}\left|\begin{array}{ccc} -2 h & -8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \\ -8 h & -512 h^{3} & 1 \end{array}\right|} \\ & =\frac{\frac{1}{2}\left|\begin{array}{ccc} h & h^{3} & 1 \\ -2 h & -8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array}\right|}{\frac{1}{2} \times(-2) \times(-8)\left|\begin{array}{ccc} h & h^{3} & 1 \\ -2 h & -8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array}\right|} \end{aligned} $$
$\Rightarrow \quad \frac{D^{\prime}}{D^{\prime}}=\frac{1}{16}=1: 16$ which is the required ratio.
