Tangent at a point $P_{1}$ (other) then $(0,0)$ on the curve $y=x^{3}$ meets the curve again at $P_{

Question

Tangent at a point $P_{1}$ (other) then $(0,0)$ 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. The abscissae of $P_{1}, P_{2}, P_{3}, \ldots \ldots . ., P_{n}$ form a G.P. The ratio $\frac{\left[\operatorname{area}\left(\Delta P_{2} P_{3} P_{4}\right)\right]}{\left[\operatorname{area}\left(\Delta P_{1} P_{2} P_{3}\right)\right]}$.

Professor PC Thomas Classes · Answer

Let any point $P_{1}$ on $y=x^{3}$ be $\left(h, h^{3}ight)$ Tangent at $P_{1}$ is\ $y-h^{3}=3 h^{2}(x-h)$\ It meets $y=x^{3}$ at $P_{2}$\ $	herefore \quad x^{3}-h^{3}=3 h^{2}(x-h)$\ $(x-h)\left(x^{2}+x h+h^{2}ight)=3 h^{2}(x-h)$\ $\Rightarrow \quad x^{2}+x h-2 h^{2}=0$\ $\Rightarrow(x-h)(x+2 h)=0$\ Either $x=h$ or $x=-2 h$ for $P_{2}$\ $	herefore \quad P_{2}$ is $\left(-2 h,-8 h^{3}ight)$\ Tangent at $P_{2}$ is $y+8 h^{3}=3(2 h)^{2}(x+2 h)$\ It meets $y=x^{3}$ at $P_{3}$\ $x^{3}+8 h^{3}=12 h^{2}(x+2 h)$\ $	herefore \quad x^{2}-2 h x-8 h^{2}=0$\ $(x-4 h)(x+2 h)=0$\ $	herefore \quad x=4 h$ for $P_{3} \quad 	herefore P_{3}$ is $\left(4 h, 64 h^{3}ight)$\ Continuing like this, we get $x=-8 h$\ for $P_{4}$ etc.\ Hence the abscissas are $h,-2 h, 4 h,-8 h$ $\qquad$\ which form a G.P.\ It can be easily shown that $D_{2}=16 D_{1}$\ $	herefore \frac{\Delta P_{1} P_{2} P_{3}}{\Delta P_{2} P_{3} P_{4}}=\frac{D_{1}}{D_{2}}=\frac{1}{16}$.\