Let `x=2t, y =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 botBA` , where A is any point on the conic. If k is the ordinate of the centroid of the `DeltaSAB` , then `lim_(t rarr1)( k)` is equal to :

Options

A. `17/18`
B. `19/18`
C. `11/18`
D. `13/18`

Answer

To solve the problem step by step, we will follow the given information and derive the required limit.

### Step 1: Identify the Conic
Given the parametric equations:
- \( x = 2t \)
- \( y = \frac{t^2}{3} \)

We can eliminate \( t \) to find the equation of the conic. From \( x = 2t \), we have \( t = \frac{x}{2} \). Substituting this into the equation for \( y \):
\[ y = \frac{(\frac{x}{2})^2}{3} = \frac{x^2}{12} \]
This represents a parabola that opens upwards.

### Step 2: Identify the Focus and Point B
The standard form of a parabola \( y = \frac{x^2}{4a} \) indicates that \( 4a = 12 \), so \( a = 3 \). The focus \( S \) of the parabola is at \( (0, a) = (0, 3) \).

Point \( B \) lies on the axis of the parabola, which is the y-axis, and we denote it as \( B(0, b) \).

### Step 3: Determine the Slope Conditions
Let \( A \) be any point on the parabola, given by \( A(2t, \frac{t^2}{3}) \).

To find the point \( B \) such that \( SA \perp AB \), we need to calculate the slopes:
- Slope of \( SA \) from \( S(0, 3) \) to \( A(2t, \frac{t^2}{3}) \):
\[ \text{slope of } SA = \frac{\frac{t^2}{3} - 3}{2t - 0} = \frac{t^2 - 9}{6t} \]

- Slope of \( AB \) from \( A(2t, \frac{t^2}{3}) \) to \( B(0, b) \):
\[ \text{slope of } AB = \frac{b - \frac{t^2}{3}}{0 - 2t} = -\frac{b - \frac{t^2}{3}}{2t} \]

Setting the product of the slopes to \(-1\) for perpendicularity:
\[ \frac{t^2 - 9}{6t} \cdot \left(-\frac{b - \frac{t^2}{3}}{2t}\right) = -1 \]

### Step 4: Solve for \( b \)
Rearranging gives:
\[ \frac{(t^2 - 9)(b - \frac{t^2}{3})}{12t^2} = 1 \]
Multiplying both sides by \( 12t^2 \):
\[ (t^2 - 9)(b - \frac{t^2}{3}) = 12t^2 \]
Expanding:
\[ b(t^2 - 9) - \frac{t^4 - 27}{3} = 12t^2 \]
Thus:
\[ b(t^2 - 9) = 12t^2 + \frac{t^4 - 27}{3} \]
\[ b = \frac{12t^2 + \frac{t^4 - 27}{3}}{t^2 - 9} \]

### Step 5: Find the Centroid and its Ordinate
The centroid \( G \) of triangle \( SAB \) has coordinates:
\[ G\left(\frac{0 + 0 + 2t}{3}, \frac{3 + b + \frac{t^2}{3}}{3}\right) \]
Thus, the ordinate \( k \) of the centroid is:
\[ k = \frac{3 + b + \frac{t^2}{3}}{3} \]

### Step 6: Take the Limit as \( t \to 1 \)
Substituting \( t = 1 \) into the expression for \( b \):
\[ b = \frac{12(1) + \frac{1 - 27}{3}}{1 - 9} = \frac{12 - \frac{26}{3}}{-8} = \frac{\frac{36 - 26}{3}}{-8} = \frac{\frac{10}{3}}{-8} = -\frac{10}{24} = -\frac{5}{12} \]

Now substituting \( b \) into \( k \):
\[ k = \frac{3 - \frac{5}{12} + \frac{1}{3}}{3} = \frac{\frac{36}{12} - \frac{5}{12} + \frac{4}{12}}{3} = \frac{\frac{35}{12}}{3} = \frac{35}{36} \]

Thus, the limit is:
\[ \lim_{t \to 1} k = \frac{35}{36} \]

### Final Answer
\[ \lim_{t \to 1} k = \frac{35}{36} \]