Let A (0,0) and B(8,2) be two fixed points on the curve `y^(3) =|x|` A point C (abscissa is less than 0) starts moving from origin along the curve such that rate of change in the ordinate is 2 cm/sec. After `t_(0)` seconds, triangle ABC becomes a right triangle.
After `t_(0)` secods, tangent is drawn to teh curve at point C to intersect it again at `(alpha,beta).` Then the value of `4alpha+3 beta` is

Options

A. `4/3`
B. `3/4`
C. 2
D. 1

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and apply the necessary mathematical concepts.

### Step 1: Understand the Curve
The curve is given by the equation:
\[ y^3 = |x| \]
Since point C has an abscissa (x-coordinate) less than 0, we can write:
\[ y^3 = -x \]

### Step 2: Rate of Change of Ordinate
The rate of change of the ordinate (y-coordinate) is given as:
\[ \frac{dy}{dt} = 2 \text{ cm/sec} \]

### Step 3: Position of Point C
Let the coordinates of point C be \( (x, y) \). Since point C is moving along the curve, we can express \( y \) in terms of \( x \):
\[ y = (-x)^{1/3} \]

### Step 4: Find the Relationship Between x and y
Using the chain rule, we can relate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
First, we need to find \( \frac{dy}{dx} \):
\[ y = (-x)^{1/3} \]
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{3}(-x)^{-2/3} \cdot (-1) = -\frac{1}{3}(-x)^{-2/3} \]

Now substituting \( \frac{dy}{dt} \):
\[ 2 = -\frac{1}{3}(-x)^{-2/3} \cdot \frac{dx}{dt} \]
This gives us:
\[ \frac{dx}{dt} = -6(-x)^{2/3} \]

### Step 5: Determine the Position After \( t_0 \) Seconds
Let’s find the position of C after \( t_0 \) seconds. We know that:
\[ y = \frac{dy}{dt} \cdot t_0 = 2t_0 \]
Thus:
\[ 2t_0 = (-x)^{1/3} \]
Cubing both sides:
\[ 8t_0^3 = -x \]
This implies:
\[ x = -8t_0^3 \]

### Step 6: Find y-coordinate
Substituting \( x \) back into the equation for \( y \):
\[ y = (-(-8t_0^3))^{1/3} = (8t_0^3)^{1/3} = 2t_0 \]

### Step 7: Slope of Line AB and BC
The coordinates of points A, B, and C are:
- A(0, 0)
- B(8, 2)
- C(-8t_0^3, 2t_0)

The slope of line AB:
\[ m_{AB} = \frac{2 - 0}{8 - 0} = \frac{1}{4} \]

The slope of line BC:
\[ m_{BC} = \frac{2t_0 - 2}{-8t_0^3 - 8} \]

### Step 8: Right Triangle Condition
For triangle ABC to be a right triangle, the product of the slopes must equal -1:
\[ m_{AB} \cdot m_{BC} = -1 \]
Substituting the slopes:
\[ \frac{1}{4} \cdot \frac{2t_0 - 2}{-8t_0^3 - 8} = -1 \]
This leads to:
\[ 2t_0 - 2 = 4(8t_0^3 + 8) \]
Expanding and simplifying gives:
\[ 2t_0 - 2 = 32t_0^3 + 32 \]
Rearranging leads to a cubic equation.

### Step 9: Find the Tangent Line
After \( t_0 \) seconds, we need to find the tangent at point C. The slope of the tangent line at C is given by \( \frac{dy}{dx} \) evaluated at C:
\[ m_{C} = -\frac{1}{3}(-(-8t_0^3))^{-2/3} = -\frac{1}{3}(8t_0^3)^{-2/3} \]

### Step 10: Intersection of Tangent with Curve
The tangent line equation can be found and set equal to the curve equation to find the coordinates \( (\alpha, \beta) \).

### Final Calculation
Finally, we need to compute \( 4\alpha + 3\beta \).