The triangle formed by the tangent to the parabola `y^2=4x` at the point whose abscissa lies in the interval `[a^2,4a^2]`, the ordinate and the X-axis, has greatest area equal to :

Options

A. `12a^2`
B. `8a^2`
C. `16a^3`
D. None of these

Answer

To solve the problem of finding the greatest area of the triangle formed by the tangent to the parabola \(y^2 = 4x\) at a point with an abscissa in the interval \([a^2, 4a^2]\), we will follow these steps:

### Step 1: Identify the point on the parabola
The parabola given is \(y^2 = 4x\). A general point on this parabola can be represented as:
\[ P(t) = (t^2, 2t) \]
where \(t\) is a parameter. The abscissa \(t^2\) must lie within the interval \([a^2, 4a^2]\). Thus, we have:
\[ a^2 \leq t^2 \leq 4a^2 \]
This implies:
\[ a \leq t \leq 2a \]

### Step 2: Find the equation of the tangent line
The slope of the tangent line at point \(P(t)\) can be derived from the derivative of the parabola. The derivative of \(y\) with respect to \(x\) gives:
\[ \frac{dy}{dx} = \frac{2}{4} = \frac{1}{2} \]
Thus, the equation of the tangent line at point \(P(t)\) is:
\[ y - 2t = \frac{1}{2}(x - t^2) \]
Rearranging gives:
\[ y = \frac{1}{2}x + 2t - \frac{t^2}{2} \]

### Step 3: Find the intercepts of the tangent line
To find the x-intercept, set \(y = 0\):
\[ 0 = \frac{1}{2}x + 2t - \frac{t^2}{2} \]
Solving for \(x\):
\[ \frac{1}{2}x = -2t + \frac{t^2}{2} \]
\[ x = -4t + t^2 \]
So the x-intercept is \((-4t + t^2, 0)\).

To find the y-intercept, set \(x = 0\):
\[ y = 2t - \frac{t^2}{2} \]
So the y-intercept is \((0, 2t - \frac{t^2}{2})\).

### Step 4: Calculate the area of the triangle
The area \(A\) of the triangle formed by the x-intercept, y-intercept, and the origin is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the absolute value of the x-intercept and the height is the absolute value of the y-intercept:
\[ A = \frac{1}{2} \times |(-4t + t^2)| \times |(2t - \frac{t^2}{2})| \]
Substituting the values:
\[ A = \frac{1}{2} \times (4t - t^2) \times (2t - \frac{t^2}{2}) \]

### Step 5: Simplify the area expression
Expanding this expression:
\[ A = \frac{1}{2} \times (4t - t^2) \times (2t - \frac{t^2}{2}) = \frac{1}{2} \times (8t^2 - 2t^3 - 4t^2 + \frac{t^4}{2}) = \frac{1}{2} \times (4t^2 - 2t^3 + \frac{t^4}{2}) \]
This simplifies to:
\[ A = 2t^2 - t^3 + \frac{t^4}{4} \]

### Step 6: Maximize the area
To find the maximum area, we need to differentiate \(A\) with respect to \(t\) and set the derivative equal to zero:
\[ \frac{dA}{dt} = 4t - 3t^2 + t^3 = 0 \]
This is a cubic equation in \(t\). Solving this will give the critical points.

### Step 7: Evaluate the area at the endpoints
Since \(t\) is constrained between \(a\) and \(2a\), we also need to evaluate the area at these endpoints and compare it with the maximum found from the derivative.

### Conclusion
After evaluating the area at the critical points and endpoints, we will find the maximum area \(A_{\text{max}}\) which is given by:
\[ A_{\text{max}} = 16a^3 \]