The coordinates of the point on the curve `x^(3)=y(x-a)^(2)` where the ordinate is minimum is

Options

A. `(3a,(27)/(4)a)`
B. `(2a,8a)`
C. `(a,0)`
D. None of these

Explanation

To find the coordinates of the point on the curve \( x^3 = y(x - a)^2 \) where the ordinate (y-coordinate) is minimum, we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^3 = y(x - a)^2 \]
We can express \( y \) in terms of \( x \):
\[ y = \frac{x^3}{(x - a)^2} \]

### Step 2: Differentiate \( y \)
To find the minimum value of \( y \), we need to differentiate \( y \) with respect to \( x \). We will use the quotient rule for differentiation:
\[ \frac{dy}{dx} = \frac{(x - a)^2 \cdot \frac{d}{dx}(x^3) - x^3 \cdot \frac{d}{dx}((x - a)^2)}{((x - a)^2)^2} \]
Calculating the derivatives:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( (x - a)^2 \) is \( 2(x - a) \).

Substituting these into the formula gives:
\[ \frac{dy}{dx} = \frac{(x - a)^2 \cdot 3x^2 - x^3 \cdot 2(x - a)}{(x - a)^4} \]

### Step 3: Simplify the derivative
Now we simplify the numerator:
\[ = \frac{3x^2(x - a)^2 - 2x^3(x - a)}{(x - a)^4} \]
Factoring out \( (x - a) \):
\[ = \frac{(x - a)(3x^2(x - a) - 2x^3)}{(x - a)^4} \]
This simplifies to:
\[ = \frac{3x^2(x - a) - 2x^3}{(x - a)^3} \]

### Step 4: Set the derivative to zero
To find the critical points, we set the numerator equal to zero:
\[ 3x^2(x - a) - 2x^3 = 0 \]
Factoring out \( x^2 \):
\[ x^2(3(x - a) - 2x) = 0 \]
This gives us:
\[ x^2 = 0 \quad \text{or} \quad 3(x - a) - 2x = 0 \]
From \( x^2 = 0 \), we get \( x = 0 \). For the second equation:
\[ 3x - 3a - 2x = 0 \implies x = 3a \]

### Step 5: Find the corresponding \( y \) value
Now we substitute \( x = 3a \) back into the equation for \( y \):
\[ y = \frac{(3a)^3}{(3a - a)^2} = \frac{27a^3}{(2a)^2} = \frac{27a^3}{4a^2} = \frac{27a}{4} \]

### Final Answer
Thus, the coordinates of the point on the curve where the ordinate is minimum are:
\[ (3a, \frac{27a}{4}) \]