The line `y=2x+c` is tangent to the parabola `y^(2)-4y-8x=4` at a point whose abscissa is `alpha`, then the ordered pair `(alpha, C)` isn

Options

A. `(-(1)/(2), 4)`
B. `(-(1)/(2), 5)`
C. `(4, 5)`
D. `(-(1)/(2), (1)/(2))`

Answer

To solve the problem, we need to find the ordered pair \((\alpha, c)\) where the line \(y = 2x + c\) is tangent to the parabola given by the equation \(y^2 - 4y - 8x = 4\) at a point whose abscissa is \(\alpha\).

### Step 1: Rewrite the parabola equation
First, we rewrite the parabola equation in a more standard form:
\[ y^2 - 4y - 8x - 4 = 0 \]

### Step 2: Differentiate the parabola
Next, we differentiate the parabola with respect to \(x\) to find the slope of the tangent line. Using implicit differentiation:
\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(4y) - \frac{d}{dx}(8x) - \frac{d}{dx}(4) = 0 \]
This gives:
\[ 2y \frac{dy}{dx} - 4 \frac{dy}{dx} - 8 = 0 \]

### Step 3: Factor out \(\frac{dy}{dx}\)
We can factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(2y - 4) = 8 \]
Thus, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{8}{2y - 4} \]

### Step 4: Set the slope equal to the line's slope
Since the line has a slope of 2, we set the derivative equal to 2:
\[ \frac{8}{2y - 4} = 2 \]

### Step 5: Solve for \(y\)
Cross-multiplying gives:
\[ 8 = 2(2y - 4) \]
Expanding this, we have:
\[ 8 = 4y - 8 \]
Adding 8 to both sides:
\[ 16 = 4y \]
Thus,
\[ y = 4 \]

### Step 6: Substitute \(y\) back into the parabola equation
Now we substitute \(y = 4\) back into the parabola equation to find \(\alpha\):
\[ (4)^2 - 4(4) - 8\alpha = 4 \]
This simplifies to:
\[ 16 - 16 - 8\alpha = 4 \]
Which simplifies to:
\[ -8\alpha = 4 \]
Thus,
\[ \alpha = -\frac{1}{2} \]

### Step 7: Find \(c\)
Now that we have \(\alpha\), we can find \(c\) using the line equation \(y = 2x + c\) at the point \((\alpha, y)\):
\[ 4 = 2(-\frac{1}{2}) + c \]
This simplifies to:
\[ 4 = -1 + c \]
Thus,
\[ c = 5 \]

### Final Answer
The ordered pair \((\alpha, c)\) is:
\[ \left(-\frac{1}{2}, 5\right) \]