Find the point(s) on the curve :
(i) `y=3x^(2)-12x+6`
(ii) `x^(2)+y^(2)-2x-3=0`
at which the tangent is parallel to x - axis.

Answer

To find the points on the given curves where the tangent is parallel to the x-axis, we need to determine where the derivative (slope) of the function equals zero. Let's solve each part step by step.

### (i) For the curve \( y = 3x^2 - 12x + 6 \):

1. **Differentiate the function**:
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - 12x + 6) \]
Using the power rule:
\[ \frac{dy}{dx} = 6x - 12 \]

2. **Set the derivative equal to zero**:
\[ 6x - 12 = 0 \]

3. **Solve for \( x \)**:
\[ 6x = 12 \implies x = 2 \]

4. **Find the corresponding \( y \) value**:
Substitute \( x = 2 \) back into the original equation:
\[ y = 3(2^2) - 12(2) + 6 \]
\[ y = 3(4) - 24 + 6 = 12 - 24 + 6 = -6 \]

5. **Point on the curve**:
The point is \( (2, -6) \).

### (ii) For the curve \( x^2 + y^2 - 2x - 3 = 0 \):

1. **Differentiate the equation implicitly**:
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(3) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} - 2 = 0 \]

2. **Rearrange to solve for \( \frac{dy}{dx} \)**:
\[ 2y \frac{dy}{dx} = 2 - 2x \]
\[ \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y} \]

3. **Set the derivative equal to zero**:
For the tangent to be parallel to the x-axis, we set:
\[ \frac{1 - x}{y} = 0 \]
This implies \( 1 - x = 0 \), so:
\[ x = 1 \]

4. **Find the corresponding \( y \) values**:
Substitute \( x = 1 \) back into the original equation:
\[ (1)^2 + y^2 - 2(1) - 3 = 0 \]
\[ 1 + y^2 - 2 - 3 = 0 \]
\[ y^2 - 4 = 0 \implies y^2 = 4 \implies y = \pm 2 \]

5. **Points on the curve**:
The points are \( (1, 2) \) and \( (1, -2) \).

### Final Answers:
- For part (i): The point is \( (2, -6) \).
- For part (ii): The points are \( (1, 2) \) and \( (1, -2) \).