The tangent PT and the normal PN to the parabola y² = 4ax at a point P on it meet its axis at points

Q: Wait, why do we use parametric coordinates $(at^2, 2at)$ instead of just $(x, y)$?

Great question! Using $(x, y)$ would force us to deal with square roots or implicit differentiation, which makes the algebra messy and prone to errors. Parametric coordinates $(at^2, 2at)$ are the 'secret weapon' of parabola problems—they reduce the geometry to simple polynomial manipulation, making the elimination of the parameter $t$ much cleaner.

Q: How did you get the coordinates of $T$ and $N$ so quickly?

It comes from the standard equations of the tangent and normal. For the tangent $ty = x + at^2$, setting $y=0$ gives $x = -at^2$, so $T = (-at^2, 0)$. For the normal $y + tx = 2at + at^3$, setting $y=0$ gives $tx = 2at + at^3$, so $x = 2a + at^2$, meaning $N = (2a + at^2, 0)$. These are standard results that you should definitely keep in your toolkit for JEE!

Q: Is the locus of the centroid always a parabola?

In this specific configuration, yes! When we eliminate the parameter $t$, we get a relationship where $k^2$ is proportional to $h$. Specifically, we found $k^2 = \frac{4a}{3}(h - \frac{2a}{3})$. This is the classic form of a parabola $Y^2 = 4AX$, just shifted. It's a beautiful example of how geometric constructions often lead to familiar algebraic curves.

Question

The tangent $PT$ and the normal $PN$ to the parabola $y^2 = 4ax$ at a point $P$ on it meet its axis at points $T$ and $N$, respectively. The locus of the centroid of the triangle $PTN$ is a parabola whose

Options:

(A) vertex is $(\frac{2a}{3}, 0)$

(B) directrix is $x = 0$

(C) latus rectum is $\frac{2a}{3}$

(D) focus is $(a, 0)$

Sigma Percentile · Accepted Answer

### The Geometry of Motion: Unveiling the Locus of the Centroid

Welcome, future engineers! Today, we are going to embark on a journey through the elegant world of coordinate geometry. We are looking at a classic JEE Advanced problem involving the parabola $y^2 = 4ax$.

It might seem like a simple curve, but when we start drawing tangents and normals, we uncover a hidden structure. Imagine you are standing on the curve of the parabola $y^2 = 4ax$. You pick a point $P$ and draw a tangent $PT$ and a normal $PN$.

These lines slice through the axis of the parabola at points $T$ and $N$. Our goal is to find the path—the locus—traced by the centroid of the triangle formed by $P$, $T$, and $N$.

### Phase 1: The Power of Parametric Coordinates

To master this problem, we must choose our tools wisely. Instead of working with arbitrary $(x, y)$ coordinates, we use the parametric form $P(at^2, 2at)$.

Because this single parameter $t$ captures the entire essence of the parabola, every point on the curve is uniquely defined by $t$. This choice transforms our geometric problem into a beautiful algebraic dance.

By using $P(at^2, 2at)$, we ensure that the point always satisfies the equation $y^2 = 4ax$ for any real value of $t$.

### Phase 2: The Intercepts $T$ and $N$

Next, we construct the tangent and the normal. The equation of the tangent at $P$ is well-known: $ty = x + at^2$.

To find where this line meets the axis of the parabola (the x-axis), we set $y=0$. This gives us $x = -at^2$, so our point $T$ is located at $(-at^2, 0)$.

Now, for the normal. The normal is perpendicular to the tangent, and its equation is $y + tx = 2at + at^3$.

Setting $y=0$ to find the intersection with the x-axis, we get $tx = 2at + at^3$, which simplifies to $x = 2a + at^2$. Thus, point $N$ is at $(2a + at^2, 0)$.

We now have our three vertices: $P(at^2, 2at)$, $T(-at^2, 0)$, and $N(2a + at^2, 0)$.

### Phase 3: The Centroid's Journey

Now, we define the centroid $G(h, k)$ of triangle $PTN$. The centroid is the 'center of mass' of the triangle, calculated as the average of the coordinates of its vertices.

For the x-coordinate $h$, we have:

$$h = \frac{at^2 + (-at^2) + (2a + at^2)}{3}$$

Notice the beautiful cancellation here! The $at^2$ and $-at^2$ terms vanish, leaving us with:

$$h = \frac{2a + at^2}{3}$$

For the y-coordinate $k$, we have:

$$k = \frac{2at + 0 + 0}{3} = \frac{2at}{3}$$

### Phase 4: Eliminating the Parameter

We are almost there. To find the locus, we need an equation relating $h$ and $k$ directly. From $k = \frac{2at}{3}$, we isolate $t$:

$$t = \frac{3k}{2a}$$

Now, we substitute this into our expression for $h$:

$$3h = 2a + a\left(\frac{3k}{2a}\right)^2$$

Expanding this, we get:

$$3h = 2a + a\left(\frac{9k^2}{4a^2}\right) \Rightarrow 3h = 2a + \frac{9k^2}{4a}$$

Rearranging to isolate $k^2$, we arrive at:

$$9k^2 = 4a(3h - 2a) \Rightarrow k^2 = \frac{4a}{3}\left(h - \frac{2a}{3}\right)$$

### Phase 5: The Final Reveal

Look at the equation $k^2 = \frac{4a}{3}(h - \frac{2a}{3})$. If we replace $h$ and $k$ with $x$ and $y$, we get:

$$y^2 = \frac{4a}{3}\left(x - \frac{2a}{3}\right)$$

This is the equation of a parabola! Comparing it to the standard form $Y^2 = 4AX$, we see the vertex is at $(\frac{2a}{3}, 0)$.

This confirms that our locus is indeed a parabola. Through this journey, we've seen how parametric coordinates and the centroid formula allow us to uncover the hidden geometry of the parabola.

Animated Solution for Mathematics - Conic Sections: The tangent $P T$ and the normal $P N$ to the parabola $y^{2} = 4 a x$ at a point $P$ on it meet its axis at points $T$ and $N$ , respectively. The locus of the centroid of the triangle $P T N$ is a parabola whose

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 4 a x$

Parametric Coordinates of $P$

Finding Point $T$ on the Axis

Finding Point $N$ on the Axis

Defining the Centroid $G (h, k)$

Calculating $h$ and $k$

Eliminating the Parameter $t$

The Locus Equation

Analyzing the New Parabola

Conclusion and Final Answer

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4ax

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

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T on the Axis

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N on the Axis

Defining the Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G(h,k)

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

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

The Locus Equation

Analyzing the New Parabola

Conclusion and Final Answer

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

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