Tangent at a point P1 {other than (0, 0)} on the curve y = x^3 meets the curve again at P2. The tang

Q: Wait, why is the root $t$ repeated twice in the cubic equation? Shouldn't it just be one root?

That is a fantastic question! When a line is tangent to a curve at a point, it doesn't just cross it; it 'kisses' the curve. Algebraically, this means the intersection point is a root of multiplicity 2. If it were a simple intersection, it would be a root of multiplicity 1. Because the tangent line shares the same slope as the curve at that point, the cubic equation $x^3 - 3t^2x + 2t^3 = 0$ must have $(x-t)^2$ as a factor. This is the geometric definition of tangency manifesting in algebra!

Q: Why does the area ratio turn out to be independent of the starting point $x_1$?

Because the curve $y=x^3$ is a homogeneous scaling structure. When you scale the x-coordinate by a factor $r$, the y-coordinate scales by $r^3$. The area, being a product of coordinates (via the determinant formula), scales by $r^4$. Since the ratio is $\frac{\Delta_1}{\Delta_2}$, the $x_1^4$ terms from the first triangle and the $x_2^4$ terms from the second triangle cancel out perfectly, leaving only the ratio of the scaling factors. It is a beautiful property of power functions!

Q: Is there a faster way to calculate the area of the triangle without using the full determinant formula?

You could use the shoelace formula, but for a curve like $y=x^3$, there is a deeper symmetry. However, the determinant method is the most robust way to prove it for any $n$. Don't fear the algebra! Once you factor out the $x_n^4$, the remaining expression is just a constant involving the common ratio $r$. The 'hard' part is just keeping your variables organized.

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. Show that the abscissae of $P_1, P_2, P_3, \dots, P_n$ form a GP. Also find the ratio $[	ext{area } (	riangle P_1P_2P_3)] / [	ext{area } (	riangle P_2P_3P_4)]$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not merely solving a problem; we are embarking on a journey to uncover the hidden rhythm of the cubic curve. Many students look at $y=x^3$ and see a simple, monotonic function.

When we introduce a tangent line, we aren't just drawing a line; we are initiating a geometric chain reaction. This problem is a classic JEE Advanced favorite because it tests your ability to bridge the gap between calculus, algebra, and geometry.

### The Tangent and the Cubic Secret

Imagine you are standing on the curve $y=x^3$ at some point $P_1(t, t^3)$. You draw a tangent line, which is destined to strike the curve again at a new point, $P_2$.

The slope of our curve is given by the derivative $\frac{dy}{dx} = 3x^2$. At our point $P_1$, the slope is $m = 3t^2$. Using the point-slope form, the equation of our tangent line is:

$$y - t^3 = 3t^2(x - t)$$

To find the intersection $P_2$, we substitute $y = x^3$ into our tangent equation:

$$x^3 - t^3 = 3t^2(x - t)$$

Rearranging this, we obtain the cubic equation:

$$x^3 - 3t^2x + 2t^3 = 0$$

Because the line is tangent at $x=t$, $t$ is a root of multiplicity two. In any cubic equation $ax^3 + bx^2 + cx + d = 0$, the sum of the roots is $-\frac{b}{a}$. Since the coefficient of $x^2$ is zero, the sum of the roots $t, t,$ and $x_2$ must be zero:

$$t + t + x_2 = 0 \Rightarrow x_2 = -2t$$

### The Geometric Progression

The next point $P_2$ has an x-coordinate that is exactly $-2$ times the previous one. If we repeat this process, the tangent at $P_2$ will hit the curve at $P_3$, where $x_3 = -2x_2$.

We have proven that the abscissae form a Geometric Progression with a common ratio $r = -2$. No matter where you start, the points will always march forward, scaling by a factor of $-2$ in the x-direction.

### The Symphony of Areas

Now, we tackle the area of the triangle formed by three consecutive points $P_n, P_{n+1}, P_{n+2}$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the determinant formula:

$$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Since our points lie on $y=x^3$, we substitute $y_i = x_i^3$. Using the GP property $x_{n+1} = rx_n$ and $x_{n+2} = r^2x_n$, we find:

$$\Delta_n = \frac{1}{2} x_n^4 |1(r^3 - r^6) + r(r^6 - 1) + r^2(1 - r^3)|$$

The term inside the absolute value is a constant $K$ that depends only on $r$. Thus, $\Delta_n = K x_n^4$. The area of the triangle is directly proportional to the fourth power of the x-coordinate of the starting point.

### The Final Act: The Ratio

We want the ratio of the areas of two consecutive triangles:

$$\frac{	ext{Area}(	riangle P_1P_2P_3)}{	ext{Area}(	riangle P_2P_3P_4)} = \frac{K x_1^4}{K x_2^4} = \left(\frac{x_1}{x_2}ight)^4$$

Since $x_2 = -2x_1$, the ratio is:

$$\left(\frac{1}{-2}ight)^4 = \frac{1}{16}$$

The final result is **$\frac{1}{16}$** (or **$0.0625$**). You have navigated the cubic curve, utilized the power of Vieta's formulas, and mastered the scaling properties of areas.

Enter Numerical Value:

Visualized Solution

$D e f inin g t h e C u r v e an d P o in t P_{1}$

$E q u a t i o n o f T an g e n t a t P_{1}$

$F in d in g I n t er sec t i o n P o in t P_{2}$

$S o l v in g f or x_{2}$

$T h e G eo m e t r i c P r o g r ess i o n$

$A r e a o f T r ian g l e Δ_{n}$

$E x p r ess in g A r e ain t er m so f x_{n} an d r$

$S im pl i f y in g t h e A r e a E x p r ess i o n$

$C a l c u l a t in g t h e A r e a R a t i o$

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Tangent and the Cubic Secret

The Geometric Progression

The Symphony of Areas

The Final Act

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheCurveandPointP1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EquationofTangentatP1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingIntersectionPointP2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solvingforx2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheGeometricProgression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AreaofTriangleΔn​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ExpressingAreaintermsofxn​andr

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheAreaExpression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingtheAreaRatio