Question

Tangent at a point \(P_{1}\) (other than \((0,0)\)) 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. Then the abscissae of \(P_{1},P_{2},P_{3},\ldots,P_{n}\) form:

• A. an A.P. with common difference 1
• B. an H.P. with common difference $\frac{1}{2}$
• C. a G.P. with common ratio 2
• D. a G.P. with common ratio $(-2)$

Hint:

For any cubic polynomial equation, the sum of the roots is fixed by the coefficient of the $x^2$ term (which is zero here). Since a tangency point counts as a double root, we can find the third intersection root instantly via: $x_1 + x_1 + x_2 = 0 \implies 2x_1 + x_2 = 0 \implies x_2 = -2x_1$!

Answer 1

The Correct Option is D

Concept: When a tangent drawn to a cubic curve $y = x^3$ at a localized point $P_1(x_1, y_1)$ intersects the curve again at a second point $P_2(x_2, y_2)$, we can establish a strict algebraic relationship between their $x$-coordinates (abscissae) by finding where the tangent line equation intersects the cubic polynomial. Step 1: Determine the equation of the tangent line at $P_1$.
Let the coordinates of the first point be $P_1(x_1, x_1^3)$. To find the slope of the tangent line, differentiate the cubic function equation with respect to $x$: \[ \frac{dy}{dx} = 3x^2 \quad \Rightarrow \quad m = 3x_1^2 \] Using the point-slope formula, write the equation of the tangent line at $P_1$: \[ y - x_1^3 = 3x_1^2(x - x_1) \quad \Rightarrow \quad y = 3x_1^2x - 3x_1^3 + x_1^3 \quad \Rightarrow \quad y = 3x_1^2x - 2x_1^3 \]

Step 2: Find the intersection points between the tangent line and the cubic curve.
To find where this tangent line meets the cubic curve $y = x^3$ again, equate their respective $y$ values: \[ x^3 = 3x_1^2x - 2x_1^3 \quad \Rightarrow \quad x^3 - 3x_1^2x + 2x_1^3 = 0 \] Since this line is tangent to the curve at $x = x_1$, the root $x = x_1$ must be a repeated root of this cubic polynomial equation. This means $(x - x_1)^2$ is a factor. Let us factor the cubic expression: \[ (x - x_1)^2(x + 2x_1) = 0 \]

Step 3: Isolate the relationship between successive abscissae.
Solving the factored polynomial yields the intersection roots: $x = x_1$ (the original tangency point) and the new intersection root: \[ x = -2x_1 \] Therefore, the abscissa of the next sequential point $P_2$ is directly given by: \[ x_2 = -2x_1 \]

Step 4: Identify the progression type formed by the sequence of roots.
Following this exact same geometric process for the next point sequence link $P_2 \rightarrow P_3$: \[ x_3 = -2x_2 = -2(-2x_1) = 4x_1 \] In general, for any subsequent point step index $n$, the coordinate satisfies the recursive sequence step: \[ x_n = -2x_{n-1} \quad \Rightarrow \quad \frac{x_n}{x_{n-1}} = -2 \] Since the ratio of any two consecutive terms in the sequence of abscissae is a fixed constant, the terms form a Geometric Progression (G.P.) with a common ratio of $-2$.