The point P of the curve `y^(2)=2x^(3)` such that the tangent at P is perpendicular to the line `4x-3y +2=0` is given by

Options

A. `(2,4)`
B. `(1,sqrt(2))`
C. `(1//2, - 1//2)`
D. `(1//8, -1//16)`

Answer

To solve the problem, we need to find the point \( P \) on the curve \( y^2 = 2x^3 \) such that the tangent at \( P \) is perpendicular to the line \( 4x - 3y + 2 = 0 \).

### Step-by-Step Solution:

1. **Find the slope of the given line**:
The equation of the line can be rewritten in slope-intercept form:
\[ 3y = 4x + 2 \implies y = \frac{4}{3}x + \frac{2}{3} \]
Thus, the slope \( m_1 \) of the line is \( \frac{4}{3} \).

2. **Determine the slope of the tangent to the curve**:
We start with the curve \( y^2 = 2x^3 \). To find the slope of the tangent, we differentiate both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 6x^2 \implies \frac{dy}{dx} = \frac{6x^2}{2y} = \frac{3x^2}{y} \]
Therefore, the slope \( m_2 \) of the tangent at point \( P(x_1, y_1) \) is:
\[ m_2 = \frac{3x_1^2}{y_1} \]

3. **Set up the condition for perpendicularity**:
Since the tangent is perpendicular to the line, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \implies \frac{4}{3} \cdot \frac{3x_1^2}{y_1} = -1 \]
Simplifying this gives:
\[ \frac{4x_1^2}{y_1} = -1 \implies 4x_1^2 = -y_1 \]

4. **Substitute \( y_1 \) from the curve equation**:
From the curve equation \( y^2 = 2x^3 \), we have:
\[ y_1 = \sqrt{2x_1^3} \quad \text{(considering the positive root for simplicity)} \]
Substitute \( y_1 \) into the equation from step 3:
\[ 4x_1^2 = -\sqrt{2x_1^3} \]
Squaring both sides to eliminate the square root gives:
\[ (4x_1^2)^2 = 2x_1^3 \implies 16x_1^4 = 2x_1^3 \]
Rearranging this, we get:
\[ 16x_1^4 - 2x_1^3 = 0 \implies 2x_1^3(8x_1 - 1) = 0 \]
This gives us two cases:
- \( 2x_1^3 = 0 \) which leads to \( x_1 = 0 \)
- \( 8x_1 - 1 = 0 \) which leads to \( x_1 = \frac{1}{8} \)

5. **Find corresponding \( y_1 \) values**:
For \( x_1 = 0 \):
\[ y_1 = \sqrt{2(0)^3} = 0 \quad \text{(Point: (0, 0))} \]
For \( x_1 = \frac{1}{8} \):
\[ y_1 = \sqrt{2\left(\frac{1}{8}\right)^3} = \sqrt{2 \cdot \frac{1}{512}} = \sqrt{\frac{1}{256}} = \frac{1}{16} \quad \text{(Point: } \left(\frac{1}{8}, \frac{1}{16}\right)\text{)} \]

6. **Conclusion**:
The points \( P \) on the curve \( y^2 = 2x^3 \) where the tangent is perpendicular to the line \( 4x - 3y + 2 = 0 \) are:
\[ (0, 0) \text{ and } \left(\frac{1}{8}, \frac{1}{16}\right) \]