Find the co-ordinates of that point on the curve`y^(2)=x^(2)(1-x)` at which the tangent drawn is perpendicular to X-axis.

Answer

To find the coordinates of the point on the curve \( y^2 = x^2(1 - x) \) at which the tangent drawn is perpendicular to the x-axis, we follow these steps:

### Step 1: Understand the condition for the tangent to be perpendicular to the x-axis
A tangent that is perpendicular to the x-axis means that its slope is undefined or infinite. This occurs when the derivative \( \frac{dy}{dx} \) is infinite.

### Step 2: Differentiate the given curve
We start with the equation of the curve:
\[ y^2 = x^2(1 - x) \]
To find \( \frac{dy}{dx} \), we will use implicit differentiation. Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = \frac{d}{dx}[x^2(1 - x)] \]
Using the product rule on the right side:
\[ \frac{d}{dx}[x^2(1 - x)] = 2x(1 - x) + x^2(-1) = 2x - 3x^2 \]
Thus, we have:
\[ 2y \frac{dy}{dx} = 2x - 3x^2 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Now, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2x - 3x^2}{2y} \]

### Step 4: Set the derivative to be undefined
For \( \frac{dy}{dx} \) to be infinite, the denominator must be zero:
\[ 2y = 0 \implies y = 0 \]

### Step 5: Substitute \( y = 0 \) back into the curve equation
Now we substitute \( y = 0 \) back into the original curve equation to find the corresponding \( x \) values:
\[ 0^2 = x^2(1 - x) \implies 0 = x^2(1 - x) \]
This equation holds true if either \( x^2 = 0 \) or \( 1 - x = 0 \):
1. \( x^2 = 0 \) gives \( x = 0 \)
2. \( 1 - x = 0 \) gives \( x = 1 \)

### Step 6: Find the coordinates
Thus, the points on the curve where the tangent is perpendicular to the x-axis are:
1. \( (0, 0) \)
2. \( (1, 0) \)

### Final Answer
The coordinates of the points are \( (0, 0) \) and \( (1, 0) \).

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