Question

Tangent at a point $P_1\left\{otherthan(0,0)\right\}$ on the curve $y=x^3$ meets the curve again at $P_2$. The tangent at $P_2$ meet the curve at $P_3$ and so on. Then the numerical value of $\frac{\left[area\left(\Delta P_1{P}_2{P}_3\right)\right]}{\left[area\left(\Delta P_2{P}_3{P}_4\right)\right]}$

• A. $1/2$
• B. $1/4$
• C. $1/8$
• D. $1/16$

Answer 1

The correct option is A $1/2$

Let any point $P_1$ on $y=x^3$ be $(h,h^3)$
Then tangent at $P_1$ is $y{-h}^3={3h}^2(x-h)$ ...(1)
It meets $y=x^3$ at $P_2$
On putting the value of $y$ in equation (1)
$x^3-h^3={3h}^2(x-h)\Rightarrow (x-h)(x^2+xh+h^2)={3h}^2(x-h)$
$\Rightarrow x^2+xh+h^2={3h}^2$ or $x=h$
$\Rightarrow x^2+xh-{2h}^2=0\Rightarrow (x-h)(x+2h)=0$
$\Rightarrow x=h$ or $x=-2h$
Therefore, $x=-2h$ is the point $P_2$ which implies $y=-{8h}^3$
Hence, point $P_2\equiv (-2h,-{8h}^2)$
Again, tangent at $P_2$ is $y+{8h}^3=3{(-2h)}^2(x+2h)$
It meets $y=x^3$ at $P_3$
$\Rightarrow x^3+{8h}^2={12h}^2(x+2h)\Rightarrow x^2-2hx-{8h}^2=0\Rightarrow (x+2h)(x-4h)=0\Rightarrow x=4h\Rightarrow y={64h}^3$
Therefore $P_3\equiv (4h,{64h}^3)$
Similarly, we get $P_4\equiv (-8h,-8^3{h}^3)$
Hence, the absissae are $h,-2h,4h,-8h,...$ which form a GP
Let $D^{\prime }=△P_1{P}_2{P}_3$ and $D"=△P_2{P}_3{P}_4$
$\frac{D^{\prime }}{D"}=\frac{△P_1{P}_2{P}_3}{△P_2{P}_3{P}_4}=\frac{\frac{1}{2}\left|\begin{array}{ccc}h& h^3& 1\\ -2h& {-8h}^3& 1\\ 4h& {64h}^3& 1\end{array}\right|}{\frac{1}{2}\left|\begin{array}{ccc}-2h& {-8h}^3& 1\\ 4h& {64h}^3& 1\\ -8k& {-512h}^3& 1\end{array}\right|}$
$=\frac{\frac{1}{2}\left|\begin{array}{ccc}h& h^3& 1\\ -2h& {-8h}^3& 1\\ 4h& {64h}^3& 1\end{array}\right|}{\frac{1}{2}\times (-2)\times (-8)\left|\begin{array}{ccc}h& h^3& 1\\ -2h& {-8h}^3& 1\\ 4h& {64h}^3& 1\end{array}\right|}$
$\Rightarrow C_2\frac{D^{\prime }}{D"}=\frac{1}{16}$ which is the required ratio