Let y² = 12x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point

Q: Why use the dot product instead of the slope product $m_1 m_2 = -1$?

You absolutely can use the slope product! It is a valid method. However, the dot product is often more robust in JEE Advanced problems because it handles vertical lines naturally without requiring special cases. It is a cleaner, more elegant way to handle orthogonality that keeps your algebraic flow uninterrupted.

Q: Why do we assume $t \neq 0$ when dividing by $-3t^2$?

Great catch! The problem defines a triangle $OPA$. If $t=0$, point $P$ coincides with the origin $O$, and the triangle collapses into a line segment, which isn't a triangle at all. We exclude this case to maintain the geometric integrity of the figure. Always check your boundary conditions!

Q: Is the locus always a parabola?

In this specific configuration, yes! It is fascinating to see how the centroid of a triangle formed by a point on a parabola and a variable point on the axis traces out another parabola. It is a beautiful symmetry of conic sections. Keep looking for these patterns; they are the fingerprints of geometry.

Question

Let $y^2 = 12x$ be the parabola with its vertex at $O$. Let $P$ be a point on the parabola and $A$ be a point on the $x$-axis such that $\angle OPA = 90^\circ$. Then the locus of the centroid of such triangles $OPA$ is :

Options:

(A) $y^2 - 2x + 8 = 0$

(B) $y^2 - 9x + 6 = 0$

(C) $y^2 - 4x + 8 = 0$

(D) $y^2 - 6x + 4 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving an equation; we are witnessing a dance.

We have a parabola, $y^2 = 12x$, and a point $P$ moving along it. As $P$ moves, it drags a triangle $OPA$ with it, forcing point $A$ to shift along the x-axis to keep the angle at $P$ perfectly square.

Our mission is to find the path—the locus—traced by the centroid of this triangle. Let us begin.

### The Power of Parametrization

First, let us ground ourselves. We are given $y^2 = 12x$. By comparing this to the standard form $y^2 = 4ax$, we immediately identify $4a = 12$, which gives us $a = 3$.

The vertex $O$ is at $(0, 0)$. Now, instead of wrestling with $x$ and $y$ as independent variables, let us use the elegance of parametric coordinates.

Any point $P$ on this parabola can be represented as $(at^2, 2at)$. Substituting our value of $a$, we get $P = (3t^2, 6t)$.

By reducing the position of $P$ to a single parameter, $t$, we have turned a two-dimensional problem into a one-dimensional journey. This is the secret to simplifying complex geometry.

### The Orthogonality Condition

The problem imposes a strict condition: $\angle OPA = 90^{\circ}$. This is the heartbeat of the problem.

Geometrically, this means the vector $\vec{PO}$ and the vector $\vec{PA}$ are perpendicular. In the language of vectors, this is a gift; it means their dot product must be zero: $\vec{PO} \cdot \vec{PA} = 0$.

Let us calculate these vectors. Since $O = (0, 0)$, the vector $\vec{PO}$ is simply the negative of the coordinates of $P$: $\vec{PO} = (-3t^2, -6t)$.

For point $A$, which lies on the x-axis, we define it as $A = (h, 0)$. Thus, $\vec{PA} = (h - 3t^2, -6t)$. Now, we perform the dot product:

$$(-3t^2)(h - 3t^2) + (-6t)(-6t) = 0$$

Expanding this, we get $-3t^2(h - 3t^2) + 36t^2 = 0$. Assuming $t 
eq 0$, we divide by $-3t^2$ to find $h - 3t^2 - 12 = 0$, which simplifies beautifully to $h = 3t^2 + 12$.

We have found the x-coordinate of $A$ in terms of our parameter $t$.

### The Centroid and the Locus

We are now ready to find the centroid $G(X, Y)$. The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is the average of its coordinates.

For our triangle $OPA$, the vertices are $O(0, 0)$, $P(3t^2, 6t)$, and $A(3t^2 + 12, 0)$. Calculating the coordinates of $G(X, Y)$:

$$X = \frac{0 + 3t^2 + (3t^2 + 12)}{3} = \frac{6t^2 + 12}{3} = 2t^2 + 4$$

$$Y = \frac{0 + 6t + 0}{3} = \frac{6t}{3} = 2t$$

We have expressed the centroid's position entirely in terms of $t$. To find the locus, we must eliminate $t$. From the equation for $Y$, we have $t = \frac{Y}{2}$.

Substituting this into the equation for $X$:

$$X = 2\left(\frac{Y}{2}ight)^2 + 4$$

### The Final Reveal

Let us simplify this expression. We have $X = 2\left(\frac{Y^2}{4}ight) + 4$, which simplifies to $X = \frac{Y^2}{2} + 4$.

Multiplying by $2$, we get $2X = Y^2 + 8$. Rearranging, we arrive at the final equation of the locus:

$$Y^2 - 2X + 8 = 0$$

Replacing $X$ and $Y$ with the standard variables $x$ and $y$, we get **$y^2 - 2x + 8 = 0$**.

Look at that! The centroid itself traces out a parabola. It is a perfect, elegant conclusion to our journey.

Animated Solution for Mathematics - Conic Sections: Let $y^{2} = 12 x$ be the parabola with its vertex at $O$ . Let $P$ be a point on the parabola and $A$ be a point on the $x$ -axis such that $∠ O P A = 9 0^{\circ}$ . Then the locus of the centroid of such triangles $O P A$ is :

Select Answer:

Visualized Solution

Visualizing the Parabola

Parametric Coordinates of $P$

Defining Point $A$ and Triangle $O P A$

The Orthogonality Condition

Setting up the Dot Product

Solving for $h$

Finding the Centroid $G (X, Y)$

Substituting Coordinates into Centroid

Simplifying the Centroid Coordinates

Eliminating the Parameter $t$

The Final Locus Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of Parametrization

The Orthogonality Condition

The Centroid and the Locus

The Final Reveal

Similar Questions

Let $A$ be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ intersect the parabola at two distinct points $B$ and $C$ . If the centroid of the triangle $A B C$ is $(\frac{7}{3}, \frac{4}{3})$ , then $(B C)^{2}$ is equal to :

If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ and a point P moves on the line $2 x - 3 y + 4 = 0$ then the centroid of $△ P A B$ lies on the line :

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Through the vertex $O$ of parabola $y^{2} = 4 x$ , chords $O P$ and $O Q$ are drawn at right angles to one another. Show that for all positions of $P$ , $P Q$ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of $P Q$ .

Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Select Answer:

Visualized Solution

Visualizing the Parabola

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

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPA

The Orthogonality Condition

Setting up the Dot Product

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

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

Substituting Coordinates into Centroid

Simplifying the Centroid Coordinates

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

The Final Locus Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of Parametrization

The Orthogonality Condition

The Centroid and the Locus

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=0, and its foci is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point (1,0) and Q a point on the locus y2=8x. The locus of mid point of PQ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

Parametric Coordinates of $P$

Defining Point $A$ and Triangle $O P A$

Solving for $h$

Finding the Centroid $G (X, Y)$

Eliminating the Parameter $t$

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is