If the tangent to the curve y = x³ at the point P (t, t³) meets the curve again at Q, then the ordin

Q: Wait, why do we divide by $(x-t)$ in the intersection step? Isn't that dangerous?

That is a brilliant observation! Dividing by $(x-t)$ is only valid if $x \neq t$. Since $P$ is the point of tangency, $x=t$ is a root of the intersection equation. Because it is a tangent, it is actually a *repeated* root. By dividing by $(x-t)$, we are effectively 'removing' the information about point $P$ to isolate the information about point $Q$.

Q: Why does the tangent to a cubic curve always hit it again at another point?

Think of the degree of the polynomial. A line is $y = mx + c$ (degree 1), and the curve is $y = x^3$ (degree 3). When you equate them, you get a cubic equation. A cubic equation must have 3 roots. At the point of tangency, the line 'touches' the curve, which counts as two roots (a double root). The third root must be the point $Q$ where the line crosses the curve again!

Q: Can I use the slope formula directly to find $Q$?

You could, but it is much more tedious. The algebraic approach of setting the line equation equal to the curve is the most robust method. It guarantees you won't miss the intersection point.

Q: Does the ratio $1:2$ mean I should use the distance formula?

Not at all! The section formula is specifically designed for coordinates. Since we are looking for the ordinate (the $y$-coordinate), we only need the $y$-values of $P$ and $Q$ and the ratio. The distance formula would introduce unnecessary square roots.

Q: Is the result $-2t^3$ always true regardless of the value of $t$?

Yes! That is the beauty of this problem. The relationship between the point of tangency and the point of intersection is a geometric property of the cubic curve $y=x^3$. It is independent of the specific value of $t$.

Question

If the tangent to the curve $y = x^3$ at the point $P (t, t^3)$ meets the curve again at $Q$, then the ordinate of the point which divides $PQ$ internally in the ratio $1: 2$ is :

Options:

(A) $-2t^3$

(B) $-t^3$

(C) $0$

(D) $2t^3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the graph of $y = x^3$. It is a smooth, elegant curve that starts from the depths of the third quadrant, passes gracefully through the origin, and climbs toward infinity in the first quadrant.

To explore the relationship between a tangent line and the point where it 'revisits' the curve, we pick an arbitrary point $P$ on the curve with coordinates $(t, t^3)$.

### The Tangent's Identity

To find the tangent line, we need its slope. We invoke the power of calculus: differentiating $y = x^3$ gives us the derivative $\frac{dy}{dx} = 3x^2$.

At our specific point $P$, the slope $m$ is simply $3t^2$. Using the point-slope form, $y - y_1 = m(x - x_1)$, we construct the equation of our tangent line:

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

This line is our probe. It captures the local behavior of the curve at $P$.

### The Intersection Mystery

Now, we ask: where does this line meet the curve again? We set the line equal to the curve: $x^3 = 3t^2(x - t) + t^3$.

Rearranging this, we get the cubic equation:

$$x^3 - t^3 - 3t^2(x - t) = 0$$

We know that $x^3 - t^3$ factors into $(x - t)(x^2 + xt + t^2)$. Substituting this back, we see:

$$(x - t)(x^2 + xt + t^2) - 3t^2(x - t) = 0$$

Since $x 
eq t$ at point $Q$, we can safely divide by $(x - t)$. This leaves us with a beautiful quadratic equation: $x^2 + xt - 2t^2 = 0$.

Factoring this, we find $(x - t)(x + 2t) = 0$. The root $x = t$ corresponds to our starting point $P$, and the root $x = -2t$ reveals the location of our mysterious point $Q$.

### The Final Synthesis

With $x_Q = -2t$, we find the ordinate $y_Q$ by plugging it back into the original curve: $y_Q = (-2t)^3 = -8t^3$. Now, we have point $P(t, t^3)$ and point $Q(-2t, -8t^3)$.

We are tasked with finding the ordinate of a point $R$ that divides $PQ$ in a $1:2$ ratio. Using the section formula, $y_R = \frac{m y_Q + n y_P}{m + n}$, we substitute our values:

$$y_R = \frac{1(-8t^3) + 2(t^3)}{1 + 2} = \frac{-8t^3 + 2t^3}{3} = \frac{-6t^3}{3} = -2t^3$$

### The Takeaway

Look at that result: **$-2t^3$**. It is elegant, simple, and profound.

We have discovered that for any tangent drawn to the curve $y = x^3$, the point of intersection $Q$ and the point of tangency $P$ are linked by a rigid geometric ratio. This is the kind of mathematical harmony that JEE Advanced problems are built upon.

Animated Solution for Mathematics - Differentiation: If the tangent to the curve $y = x^{3}$ at the point $P (t, t^{3})$ meets the curve again at $Q$ , then the ordinate of the point which divides $P Q$ internally in the ratio $1 : 2$ is :

Select Answer:

Visualized Solution

Visualizing the Curve $y = x^{3}$

Finding the Slope of Tangent

Equation of the Tangent Line

Intersection with the Curve

Factoring the Equation

Solving for $x$ -coordinate of $Q$

Finding the Ordinate of $Q$

The Section Formula

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Tangent's Identity

The Intersection Mystery

The Final Synthesis

The Takeaway

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