A point on the curve `y^(2) = 18 x` at which the ordinate increases at twice the rate of abscissa is

Options

A. (2,4)
B. (2,-4)
C. `(-(9)/(8),(9)/(2))`
D. `((9)/(8),(9)/(2))`

Answer

To solve the problem, we need to find a point on the curve defined by the equation \( y^2 = 18x \) where the ordinate (y-coordinate) increases at twice the rate of the abscissa (x-coordinate).

### Step-by-Step Solution:

1. **Differentiate the Curve Equation**:
We start with the equation of the curve:
\[ y^2 = 18x \]
To find the relationship between \( y \) and \( x \) with respect to time, we differentiate both sides with respect to \( t \):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(18x) \]
Using the chain rule, we have:
\[ 2y \frac{dy}{dt} = 18 \frac{dx}{dt} \]

2. **Apply the Given Condition**:
We are given that the ordinate increases at twice the rate of the abscissa, which means:
\[ \frac{dy}{dt} = 2 \frac{dx}{dt} \]
Substitute this into the differentiated equation:
\[ 2y(2 \frac{dx}{dt}) = 18 \frac{dx}{dt} \]
Simplifying this, we get:
\[ 4y \frac{dx}{dt} = 18 \frac{dx}{dt} \]

3. **Cancel \( \frac{dx}{dt} \)**:
Assuming \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( \frac{dx}{dt} \):
\[ 4y = 18 \]
Thus, we find:
\[ y = \frac{18}{4} = \frac{9}{2} \]

4. **Find the Corresponding \( x \)-Coordinate**:
Now that we have \( y = \frac{9}{2} \), we substitute this back into the original curve equation to find \( x \):
\[ \left(\frac{9}{2}\right)^2 = 18x \]
Calculating \( \left(\frac{9}{2}\right)^2 \):
\[ \frac{81}{4} = 18x \]
Solving for \( x \):
\[ x = \frac{81}{4 \cdot 18} = \frac{81}{72} = \frac{9}{8} \]

5. **Final Point**:
Thus, the point on the curve is:
\[ \left(\frac{9}{8}, \frac{9}{2}\right) \]

### Conclusion:
The point on the curve \( y^2 = 18x \) at which the ordinate increases at twice the rate of the abscissa is \( \left(\frac{9}{8}, \frac{9}{2}\right) \).