Find the point on the parabola `y^(2)=18x` at which ordinate is 3 times its abscissa.

Answer

To find the point on the parabola \( y^2 = 18x \) at which the ordinate (y-coordinate) is three times its abscissa (x-coordinate), we can follow these steps:

### Step 1: Understand the given parabola
The equation of the parabola is given by:
\[ y^2 = 18x \]
This can be compared with the standard form of a parabola \( y^2 = 4ax \). Here, we can identify \( 4a = 18 \).

### Step 2: Find the value of \( a \)
To find \( a \), we solve:
\[ 4a = 18 \implies a = \frac{18}{4} = \frac{9}{2} \]

### Step 3: Use parametric coordinates
For the parabola \( y^2 = 4ax \), the parametric coordinates are given by:
\[ (x, y) = (at^2, 2at) \]
Substituting \( a = \frac{9}{2} \):
\[ (x, y) = \left(\frac{9}{2}t^2, 9t\right) \]

### Step 4: Set up the condition for the ordinate being three times the abscissa
According to the problem, the ordinate \( y \) is three times the abscissa \( x \):
\[ y = 3x \]
Substituting the parametric equations:
\[ 9t = 3\left(\frac{9}{2}t^2\right) \]

### Step 5: Simplify the equation
This simplifies to:
\[ 9t = \frac{27}{2}t^2 \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 18t = 27t^2 \]
Rearranging gives:
\[ 27t^2 - 18t = 0 \]

### Step 6: Factor the equation
Factoring out \( t \):
\[ t(27t - 18) = 0 \]
This gives us two solutions:
\[ t = 0 \quad \text{or} \quad 27t - 18 = 0 \implies t = \frac{18}{27} = \frac{2}{3} \]

### Step 7: Find the points corresponding to the values of \( t \)
1. For \( t = 0 \):
\[ (x, y) = \left(\frac{9}{2}(0)^2, 9(0)\right) = (0, 0) \]

2. For \( t = \frac{2}{3} \):
\[ x = \frac{9}{2}\left(\frac{2}{3}\right)^2 = \frac{9}{2} \cdot \frac{4}{9} = 2 \]
\[ y = 9\left(\frac{2}{3}\right) = 6 \]
Thus, the point is \( (2, 6) \).

### Final Answer
The points on the parabola where the ordinate is three times the abscissa are:
\[ (0, 0) \quad \text{and} \quad (2, 6) \]