The normal at the point (bt₁², 2bt₁) on a parabola meets the parabola again in the point (bt₂², 2bt₂

Q: Wait, why can we just cancel out $(t_2 - t_1)$? Isn't that dangerous?

That is a fantastic question! In algebra, you can only divide by a term if you are certain it is not zero. Here, $t_2 - t_1 = 0$ would imply $t_2 = t_1$, which means points $P$ and $Q$ are the same point. But the problem states the normal meets the parabola *again* at $Q$, implying $P$ and $Q$ are distinct. Therefore, $t_2 \neq t_1$, and $(t_2 - t_1)$ is guaranteed to be non-zero. You are safe to cancel!

Q: Why is the normal equation $y = -tx + 2bt + bt^3$?

Great catch. The slope of the tangent at $t$ is $1/t$. Since the normal is perpendicular, its slope is $-t$. Using the point-slope form with point $(bt^2, 2bt)$, we get $y - 2bt = -t(x - bt^2)$. Expanding this gives $y - 2bt = -tx + bt^3$, which rearranges to $y = -tx + 2bt + bt^3$. It's a standard result, but deriving it once is the best way to remember it!

Q: What if $t_1 = 0$? Does the formula still work?

If $t_1 = 0$, point $P$ is the vertex $(0,0)$. The normal at the vertex is the x-axis ($y=0$). The x-axis intersects the parabola only at the vertex itself. In our formula $t_2 = -t_1 - 2/t_1$, if $t_1 = 0$, we get a division by zero, which confirms that the normal at the vertex does not intersect the parabola at a *second* distinct point. The math perfectly reflects the geometry!

Question

The normal at the point $(bt_1^2, 2bt_1)$ on a parabola meets the parabola again in the point $(bt_2^2, 2bt_2)$, then

Options:

(A) $t_2 = t_1 + \frac{2}{t_1}$

(B) $t_2 = -t_1 - \frac{2}{t_1}$

(C) $t_2 = -t_1 + \frac{2}{t_1}$

(D) $t_2 = t_1 - \frac{2}{t_1}$

Sigma Percentile · Accepted Answer

### The Geometric Dance of the Normal

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are witnessing a beautiful, elegant dance between algebra and geometry.

We are looking at a parabola, that iconic curve defined by $y^2 = 4bx$, and we are exploring the behavior of its normal lines. Imagine you are standing on this curve at a point $P$. You draw a tangent, and then, with a sharp turn, you draw a line perfectly perpendicular to it—the normal.

This line cuts through the interior of the parabola and, like a boomerang, strikes the curve again at a new point $Q$. Our mission is to find the secret relationship between the parameters of these two points, $t_1$ and $t_2$.

### Phase 1: Defining the Players

Let us ground ourselves. We define our parabola as $y^2 = 4bx$. Any point on this curve can be described by a single parameter $t$.

For our starting point $P$, we choose the parameter $t_1$, giving us the coordinates $(bt_1^2, 2bt_1)$. This is our anchor.

Now, consider the normal line at $P$. The standard equation for a normal to this parabola at any parameter $t$ is given by the elegant expression $y = -tx + 2bt + bt^3$. By substituting our specific parameter $t_1$, the equation of our normal line becomes:

$$y = -t_1x + 2bt_1 + bt_1^3$$

This line is our path. It starts at $P$ and travels across the parabola to find its second home at $Q$.

### Phase 2: The Intersection

Now, let us turn our attention to point $Q$. Since $Q$ also lies on the parabola, it must have its own parameter, $t_2$, and thus its coordinates are $(bt_2^2, 2bt_2)$.

The magic happens here: because the normal line passes through $Q$, the coordinates of $Q$ must satisfy the equation of the normal. This is the moment where geometry yields to algebra. We substitute $x = bt_2^2$ and $y = 2bt_2$ into our normal equation:

$$2bt_2 = -t_1(bt_2^2) + 2bt_1 + bt_1^3$$

Look at this equation. It looks a bit cluttered, but notice the constant $b$ in every single term. Since $b$ is the scale factor of our parabola and is non-zero, we can divide the entire equation by $b$. This simple act of cleaning up reveals the structure beneath:

$$2t_2 = -t_1t_2^2 + 2t_1 + t_1^3$$

### Phase 3: The Algebraic Unveiling

We are close. Our goal is to find the relationship between $t_1$ and $t_2$. Let us group the terms to see if we can factorize. We move the linear terms to one side:

$$2t_2 - 2t_1 = -t_1t_2^2 + t_1^3$$

On the left, we factor out a $2$, giving us $2(t_2 - t_1)$. On the right, we factor out a $-t_1$, which leaves us with $(t_2^2 - t_1^2)$. The equation now stands as:

$$2(t_2 - t_1) = -t_1(t_2^2 - t_1^2)$$

Do you see it? The term $(t_2^2 - t_1^2)$ is a classic difference of squares, which expands to $(t_2 - t_1)(t_2 + t_1)$. Substituting this back, we get:

$$2(t_2 - t_1) = -t_1(t_2 - t_1)(t_2 + t_1)$$

### The Grand Finale

Here is the moment of truth. We have a common factor of $(t_2 - t_1)$ on both sides. Because $P$ and $Q$ are distinct points, $t_2 
eq t_1$, meaning $(t_2 - t_1) 
eq 0$.

We can safely divide both sides by this factor. The equation collapses into something remarkably simple:

$$2 = -t_1(t_2 + t_1)$$

Now, we just need to isolate $t_2$. Dividing by $-t_1$, we get $t_2 + t_1 = -\frac{2}{t_1}$. Finally, subtracting $t_1$ from both sides, we arrive at our destination:

$$t_2 = -t_1 - \frac{2}{t_1}$$

This is the beautiful, universal condition for the normal at $t_1$ to intersect the parabola again at $t_2$. It is a result that appears simple, but it encapsulates the entire geometric journey we just took.

**Remember this result**, but more importantly, remember the process—the way we translated geometry into algebra and simplified it until the truth revealed itself. Keep practicing, keep questioning, and keep falling in love with the physics and math behind these problems.

Animated Solution for Mathematics - Conic Sections: The normal at the point $(b t_{1}^{2}, 2 b t_{1})$ on a parabola meets the parabola again in the point $(b t_{2}^{2}, 2 b t_{2})$ , then

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Visualized Solution

Setting up the Parabola

The Normal at Point $P$

Intersection at Point $Q$

Equation of the Normal

Substituting $Q$ into the Normal

Simplifying the Equation

Grouping Similar Terms

Factoring the Expression

Expanding the Difference of Squares

Handling the Common Factor

Canceling the Non-Zero Factor

Isolating the Parameters

Final Relation for $t_{2}$

The Sigma Insight: Equation of Tangent and Normal

The Geometric Dance of the Normal

Phase 1

Phase 2

Phase 3

The Grand Finale

Similar Questions

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Comprehension Passage

The value of $r$ is

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal at the point (bt12​,2bt1​) on a parabola meets the parabola again in the point (bt22​,2bt2​), then