Let x = 2t, y = t²/3 be a conic. Let S be the focus and B be the point on the axis of the conic such

Q: Wait, why do we assume $B$ is on the $y$-axis? The problem says 'axis of the conic', but how do I know which one it is?

Great question! By eliminating the parameter $t$ from $x = 2t$ and $y = \frac{t^2}{3}$, we get $x^2 = 12y$. This is a standard parabola opening upwards. Its axis of symmetry is the line where the $x$-term vanishes, which is the $y$-axis ($x=0$). Always reduce parametric forms to Cartesian first to see the 'skeleton' of the curve!

Q: Can I use the property that the product of slopes of perpendicular lines is $-1$ even if one line is vertical?

You are absolutely right to be cautious! If a line were vertical, its slope would be undefined. However, here $S=(0,3)$ and $A=(2t, \frac{t^2}{3})$. For $SA$ to be vertical, $2t$ would have to be $0$, meaning $t=0$. At $t=0$, $A$ is at the origin $(0,0)$. The slope calculation holds for all $t \neq 0$, and the limit as $t \to 1$ is perfectly safe.

Q: Why did we substitute $3\alpha$ instead of solving for $\alpha$ first?

It is a tactical choice! Notice that the formula for the centroid $k$ involves $\alpha$ as part of the sum $y_S + y_A + y_B$. Since $y_B = \alpha$, we have $3\alpha$ appearing naturally in the numerator. Keeping it as $3\alpha$ saves us from dividing by 3 and then multiplying by 3 again, reducing the chance of arithmetic errors.

Q: Is it possible that $SA$ and $BA$ are the same line?

If they were the same line, the angle between them would be $0$, not $90$ degrees. The condition $SA \perp BA$ forces them to be distinct lines intersecting at $A$. The algebra naturally handles this—if the slopes were equal, the product would be positive, not $-1$.

Q: Does the limit $\lim_{t \to 1} k$ imply that the centroid is moving?

Exactly! As $t$ changes, point $A$ slides along the parabola, and point $B$ must shift along the $y$-axis to maintain the perpendicularity condition. Consequently, the centroid $G$ traces a path. We are simply 'freezing' the frame at the exact moment $t=1$ to find the height of the centroid at that specific instant.

Question

Let $x = 2t, y = \frac{t^2}{3}$ be a conic. Let $S$ be the focus and $B$ be the point on the axis of the conic such that $SA \perp BA$, where $A$ is any point on the conic. If $k$ is the ordinate of the centroid of $\Delta SAB$, then $\lim_{t 	o 1} k$ is equal to

Options:

(A) $\frac{17}{18}$

(B) $\frac{19}{18}$

(C) $\frac{11}{18}$

(D) $\frac{13}{18}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, watching a point $A$ dance along a path defined by $x = 2t$ and $y = \frac{t^2}{3}$. By eliminating the parameter $t$, we reveal the true identity of this path: $x^2 = 12y$.

This is a parabola, an upward-opening curve with its vertex at the origin. Its focus $S$ sits gracefully at $(0, 3)$.

### The Perpendicularity Constraint

Now, introduce a second player: point $B$. We are told $B$ resides on the axis of the conic—the $y$-axis—so its coordinates must be $(0, \alpha)$.

The problem imposes a strict geometric law: for any position of $A$, the line segment $SA$ must be perpendicular to $BA$. As $A$ moves, $B$ must adjust its position on the $y$-axis to keep the angle $\angle SAB$ at exactly $90^\circ$.

To capture this mathematically, we invoke the condition of perpendicularity: the product of the slopes of two lines must be $-1$. The slope of $SA$ is given by:

$$m_{SA} = \frac{\frac{t^2}{3} - 3}{2t - 0} = \frac{t^2 - 9}{6t}$$

Similarly, the slope of $BA$ is:

$$m_{BA} = \frac{\frac{t^2}{3} - \alpha}{2t - 0} = \frac{t^2 - 3\alpha}{6t}$$

Setting their product to $-1$ gives us the master equation:

$$\left(\frac{t^2 - 9}{6t}ight) 	imes \left(\frac{t^2 - 3\alpha}{6t}ight) = -1$$

### Solving for the Unknown

This equation is the heartbeat of our problem. By rearranging the terms, we isolate the unknown ordinate $\alpha$ of point $B$.

Multiplying through, we find that $(t^2 - 9)(t^2 - 3\alpha) = -36t^2$. With a bit of algebraic finesse, we solve for $3\alpha$:

$$3\alpha = \frac{t^4 + 27t^2}{t^2 - 9}$$

### The Centroid's Journey

Finally, we turn our attention to the centroid $G$ of $\Delta SAB$. The centroid's ordinate $k$ is simply the average of the $y$-coordinates of the vertices $S, A,$ and $B$.

Thus, $k = \frac{y_S + y_A + y_B}{3}$. Substituting our known values, we get:

$$k = \frac{3 + \frac{t^2}{3} + \alpha}{3} = \frac{9 + t^2 + 3\alpha}{9}$$

Substituting our expression for $3\alpha$ into this formula, we obtain a single, elegant function of $t$ that describes the height of the centroid.

### The Final Limit

As we take the limit $\lim_{t 	o 1} k$, we substitute $t = 1$ into our expression:

$$k = \frac{9 + 1^2 + \frac{1^4 + 27(1)^2}{1^2 - 9}}{9} = \frac{10 + \frac{28}{-8}}{9}$$

Simplifying the fraction $\frac{28}{-8}$ to $-\frac{7}{2}$, we find:

$$k = \frac{10 - 3.5}{9} = \frac{6.5}{9} = \frac{13}{18}$$

Through the interplay of coordinate geometry and limits, we have tracked the centroid to its destination. The final value is **$\frac{13}{18}$**.

Select Answer:

Visualized Solution

Identifying the Conic Section

Locating the Focus $S$ and Axis

Defining Points $A$ and $B$

Applying Perpendicularity Condition

Setting Up the Slope Equation

Solving for $α$

Finding the Centroid's Ordinate $k$

Substituting $3 α$ into $k$

Evaluating the Limit as $t \to 1$

Final Calculation

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

Analyzing the Setup

The Perpendicularity Constraint

Solving for the Unknown

The Centroid's Journey

The Final Limit

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