Question

Tangent at a point ${\mathrm{P}}_1$, other than $\left(0,0\right)$ on the curve $\mathrm{y}={\mathrm{x}}^3$ meets the curve again at ${\mathrm{P}}_2$. The tangent at ${\mathrm{P}}_2$ meets the curve at ${\mathrm{P}}_3$ and so on. Then the abscissa of ${\mathrm{P}}_1$, ${\mathrm{P}}_2$, ${\mathrm{P}}_3$… are in:

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is B

G.P.

Explanation for correct option

Finding if the abscissae are in GP

The given curve is $y=x^3$

Then, $\frac{dy}{dx}=3x^2$

Let the coordinates of the point $P_1$be $\left(x_1,y_1\right)$

Then the slope of the tangent to the curve at

$$\begin{gathered} P_1\left(x_1,y_1\right)={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)} \\ =3{x_1}^2 \end{gathered}$$

So the equation of the tangent to the curve at $P_1$is

$$\begin{gathered} \Rightarrow y-y_1=3{x_1}^2\left(x-x_1\right) \\ \Rightarrow x^3-{x_1}^3-3{x_1}^2(x-x_1)=0 \\ \Rightarrow \left(x-x_1\right)\left(x^2+x{x}_1+{x_1}^2\right)-3{x_1}^2(x-x_1)=0 \\ \Rightarrow \left(x-x_1\right)\left(x^2+x{x}_1+{x_1}^2-3{x_1}^2\right)=0 \\ \Rightarrow \left(x-x_1\right)\left(x^2+x{x}_1-2{x_1}^2\right)=0 \\ \Rightarrow \left(x-x_1\right)\left(x^2-x{x}_1+2x{x}_1-2{x_1}^2\right)=0 \\ \Rightarrow \left(x-x_1\right)\left[x\left(x-x_1\right)+2x_1\left(x-x_1\right)\right]=0 \\ \Rightarrow \left(x-x_1\right)\left(x-x_1\right)\left(x+2x_1\right)=0 \\ \Rightarrow x=x_1\text{or}x=-2x_1 \\ \Rightarrow x=-2x_1. \end{gathered}$$

Now,

Let the coordinates of the point $P_2$be $\left(x_2,y_2\right)$

So,

$$\begin{gathered} x_2=-2x_1, \\ {y}_2={x_2}^3 \\ ={\left(-2x_1\right)}^3 \\ =-8{x_1}^3 \end{gathered}$$

This shows that the tangent at the point $P_1\left(x_1,y_1\right)$ meets the curve at $P_2\left(2x_1,-8{x_1}^3\right).$

Again the slope of the tangent to the curve at

$$\begin{gathered} P_2={\left(\frac{dy}{dx}\right)}_{\left(x_2,y_2\right)} \\ =3{x_2}^2 \end{gathered}$$

The equation of a tangent to the curve at $P_2$ is

$$\begin{gathered} \Rightarrow y-y_2=3{x_2}^2\left(x-x_2\right) \\ \Rightarrow \left(x-x_2\right)\left(x-x_2\right)\left(x+2x_2\right)=0 \\ \Rightarrow x=-2x_2. \end{gathered}$$

since,

$$\begin{gathered} x_3=-2x_2 \\ =-2\times \left(-2x_1\right) \\ =4x_1 \\ \text{and} \\ {y}_3={x_3}^3 \\ ={\left(4x_1\right)}^3 \\ =64{x_1}^3. \\ \text{So,} \\ {P}_3=\left(x_3,y_3\right) \\ =\left(4x_1,64{x_1}^3\right) \end{gathered}$$

Hence the abscissae are $x_1,-2x_1,4x_1,...$

Therefore, the abscissae are in GP in the common ratio $-2$

Therefore, the correct answer is option B.