Question

Tangent at a point $P_1$ (other than origin) 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. Then which of the following is/are correct?

• A. Abscissa of $P_1,P_2,\cdots ,P_3$ are in G.P.
  
• B. Ordinate of $P_1,P_2,\cdots ,P_3$ are in G.P.
• C. Ratio of area of $\Delta P_1{P}_2{P}_3$ and $\Delta P_2{P}_3{P}_4$ is $\frac{1}{8}$
• D. Ratio of area of $\Delta P_1{P}_2{P}_3$ and $\Delta P_2{P}_3{P}_4$ is $\frac{1}{16}$

Answer 1

Answer: (A), (B), (D)

Ratio of area of $\Delta P_1{P}_2{P}_3$ and $\Delta P_2{P}_3{P}_4$ is $\frac{1}{16}$

Let $P_1$ be $(h,h^3)$ on $y=x^3$
$\frac{dy}{dx}=3x^2$
Tangent at $P_1$ is,
$y-h^3=3h^2(x-h)\cdots \left(1\right)$
It meets again $y=x^3$ at $P_2.$
Putting the value of $y$ in eqn$\left(1\right)$
$x^3-h^3=3h^2(x-h)$
$\Rightarrow (x-h)(x^2+xh+h^2)-3h^2(x-h)=0$
$\Rightarrow (x-h{)}^2(x+2h)=0\Rightarrow x=h,-2h$
$\Rightarrow x=-2h$ is the point $P_2.$
$\Rightarrow y=-8h^3$
$\therefore P_2\equiv (-2h,-8h^3)$
Again, tangent at $P_2$ is,
$y+8h^3=3(-2h{)}^2(x+2h)$
It meets again $y=x^3$ at $P_3.$
$\Rightarrow x^3+8h^3=12h^2(x+2h)\Rightarrow (x+2h)(x^2-2hx+4h^2)-12h^2(x+2h)=0\Rightarrow (x+2h{)}^2(x-4h)=0$
$\Rightarrow x=4h,y=64h^3$
$\therefore P_3\equiv (4h,64h^3)$
$\text{Similarly,}{P}_4\equiv (-2^{3}h,-8^3{h}^3)$
Therefore, the abscissa $h,-2h,4h,-8h\cdots$ form a G.P.
The ordinate $h^3,-8h,8^{2}h,-8^{3}h\cdots$ form a G.P.

Let $T_1=\Delta P_1{P}_2{P}_3$ and $T_2=\Delta P_2{P}_3{P}_4$

$\frac{T_1}{T_2}=\frac{\Delta P_1{P}_2{P}_3}{\Delta P_2{P}_3{P}_4}=\frac{\frac{1}{2}\left|\begin{array}{ccc}h& h^3& 1\\ -2h& -8h^3& 1\\ 4h& 8^2{h}^3& 1\end{array}\right|}{\frac{1}{2}\left|\begin{array}{ccc}-2h& -8h^3& 1\\ 4h& 8^2{h}^3& 1\\ -8h& -8^3{h}^3& 1\end{array}\right|}$
$=\frac{\left|\begin{array}{ccc}h& h^3& 1\\ -2h& -8h^3& 1\\ 4h& 8^2{h}^3& 1\end{array}\right|}{(-2)(-8)\left|\begin{array}{ccc}h& h^3& 1\\ -2h& -8h^3& 1\\ 4h& 8^2{h}^3& 1\end{array}\right|}$
$=\frac{1}{16}$