Find the inclination from X-axis of the tangent drawn of the following curves at the given points:
(i) Curve`x^(2)-2y^(2)=8` at point (4, 2)
(ii) Curve `y = (x-1)(x-2)` at point (2, 0)
(iii) Curve `y^(2)=2x^(3)` at point (2, 4)

Answer

To find the inclination from the X-axis of the tangent drawn to the given curves at specified points, we will follow these steps for each part of the question.

### Part (i): Curve \(x^2 - 2y^2 = 8\) at point (4, 2)

1. **Differentiate the equation implicitly**:
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(2y^2) = \frac{d}{dx}(8) \]
This gives:
\[ 2x - 4y \frac{dy}{dx} = 0 \]

2. **Solve for \(\frac{dy}{dx}\)**:
\[ 4y \frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{2x}{4y} = \frac{x}{2y} \]

3. **Substitute the point (4, 2)**:
\[ \frac{dy}{dx} = \frac{4}{2 \cdot 2} = \frac{4}{4} = 1 \]

4. **Find the angle of inclination \(\theta\)**:
\[ \tan \theta = \frac{dy}{dx} = 1 \implies \theta = \tan^{-1}(1) = 45^\circ \]

### Part (ii): Curve \(y = (x-1)(x-2)\) at point (2, 0)

1. **Differentiate the equation**:
\[ y = x^2 - 3x + 2 \implies \frac{dy}{dx} = 2x - 3 \]

2. **Substitute the point (2, 0)**:
\[ \frac{dy}{dx} = 2 \cdot 2 - 3 = 4 - 3 = 1 \]

3. **Find the angle of inclination \(\theta\)**:
\[ \tan \theta = 1 \implies \theta = \tan^{-1}(1) = 45^\circ \]

### Part (iii): Curve \(y^2 = 2x^3\) at point (2, 4)

1. **Differentiate the equation implicitly**:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(2x^3) \implies 2y \frac{dy}{dx} = 6x^2 \]

2. **Solve for \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{6x^2}{2y} = \frac{3x^2}{y} \]

3. **Substitute the point (2, 4)**:
\[ \frac{dy}{dx} = \frac{3 \cdot (2^2)}{4} = \frac{3 \cdot 4}{4} = 3 \]

4. **Find the angle of inclination \(\theta\)**:
\[ \tan \theta = 3 \implies \theta = \tan^{-1}(3) \]

### Summary of Results:
- (i) \( \theta = 45^\circ \)
- (ii) \( \theta = 45^\circ \)
- (iii) \( \theta = \tan^{-1}(3) \)