Application of Derivatives 1 Question 19
####19. Tangent at a point $P_{1}$ {other than $\left.(0,0)\right}$ 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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Answer:
($1:16$)
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.
