To solve the problem step by step, we will analyze the given information and apply the concepts of derivatives and geometry.
### Step 1: Understand the curve and the movement of point C
The curve is given by the equation \( y^3 = |x| \). Since point C has an abscissa (x-coordinate) less than 0, we can rewrite the equation as:
\[ y^3 = -x \]
This implies:
\[ x = -y^3 \]
### Step 2: Determine the relationship between x and y
Since point C is moving along the curve, we can express the coordinates of point C as:
\[ C(-y^3, y) \]
where \( y \) is the ordinate (y-coordinate) of point C.
### Step 3: Find the rate of change of y
We are given that the rate of change of the ordinate \( y \) is \( \frac{dy}{dt} = 2 \) cm/sec.
### Step 4: Differentiate the curve equation
To find \( \frac{dx}{dt} \), we differentiate the equation \( y^3 = -x \) with respect to time \( t \):
\[
\frac{d}{dt}(y^3) = \frac{d}{dt}(-x)
\]
Using the chain rule:
\[
3y^2 \frac{dy}{dt} = -\frac{dx}{dt}
\]
Substituting \( \frac{dy}{dt} = 2 \):
\[
3y^2 \cdot 2 = -\frac{dx}{dt}
\]
Thus,
\[
\frac{dx}{dt} = -6y^2
\]
### Step 5: Find the coordinates of point C after \( t_0 \) seconds
At \( t = t_0 \), the ordinate \( y \) can be expressed as:
\[
y = 2t_0
\]
Substituting this back into the equation for \( x \):
\[
x = -y^3 = -(2t_0)^3 = -8t_0^3
\]
So, the coordinates of point C are:
\[
C(-8t_0^3, 2t_0)
\]
### Step 6: Determine when triangle ABC is a right triangle
For triangle ABC to be a right triangle, we can use the condition that the dot product of vectors AB and AC must be zero. The coordinates of points A and B are \( A(0,0) \) and \( B(8,2) \).
The vectors are:
\[
\overrightarrow{AB} = (8 - 0, 2 - 0) = (8, 2)
\]
\[
\overrightarrow{AC} = (-8t_0^3 - 0, 2t_0 - 0) = (-8t_0^3, 2t_0)
\]
The dot product is:
\[
\overrightarrow{AB} \cdot \overrightarrow{AC} = 8(-8t_0^3) + 2(2t_0) = -64t_0^3 + 4t_0
\]
Setting this equal to zero for the right triangle condition:
\[
-64t_0^3 + 4t_0 = 0
\]
Factoring out \( 4t_0 \):
\[
4t_0(-16t_0^2 + 1) = 0
\]
This gives us:
\[
t_0 = 0 \quad \text{or} \quad 16t_0^2 = 1 \Rightarrow t_0 = \frac{1}{4}
\]
### Step 7: Find the tangent at point C
Now, we need to find the tangent line at point C after \( t_0 \) seconds. The coordinates of C are:
\[
C(-8(\frac{1}{4})^3, 2(\frac{1}{4})) = C(-\frac{1}{8}, \frac{1}{2})
\]
The slope of the tangent line at C can be found using:
\[
\frac{dy}{dx} = \frac{dx/dt}{dy/dt} = \frac{-6y^2}{2} = -3y^2
\]
At \( y = \frac{1}{2} \):
\[
\frac{dy}{dx} = -3\left(\frac{1}{2}\right)^2 = -\frac{3}{4}
\]
### Step 8: Equation of the tangent line
Using point-slope form:
\[
y - \frac{1}{2} = -\frac{3}{4}\left(x + \frac{1}{8}\right)
\]
Simplifying this gives:
\[
y = -\frac{3}{4}x - \frac{3}{32} + \frac{1}{2} = -\frac{3}{4}x + \frac{16 - 3}{32} = -\frac{3}{4}x + \frac{13}{32}
\]
### Step 9: Find the intersection point (α, β)
To find the intersection of the tangent line with the curve \( y^3 = -x \), substitute \( y = -\frac{3}{4}x + \frac{13}{32} \) into \( y^3 = -x \):
Let \( y = \beta \):
\[
\beta^3 = -\alpha
\]
Substituting \( \beta = -\frac{3}{4}\alpha + \frac{13}{32} \) into this equation and solving will yield the values of \( \alpha \) and \( \beta \).
### Step 10: Calculate \( 4\alpha + 3\beta \)
After finding \( \alpha \) and \( \beta \), compute \( 4\alpha + 3\beta \).
