Find the maximum area of an isosceles triangle inscribed in the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`with its vertex at one end of the major axis.

Options

A. `sqrt(3) ab`
B. `(3sqrt3)/(4) ab`
C. `(5sqrt3)/(4)`
D. none of these

Answer

To find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with its vertex at one end of the major axis, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
The ellipse is symmetric about both axes, with its major axis along the x-axis. The vertices of the triangle will be at points \(A\), \(B\), and \(C\), where point \(A\) is at one end of the major axis, specifically at \((a, 0)\).

### Step 2: Define the Coordinates of Points \(B\) and \(C\)
Let the coordinates of point \(B\) be \((x_B, y_B)\) and point \(C\) be \((x_C, y_C)\). Since the triangle is isosceles and symmetric about the y-axis, we can define:
- \(B = (a \cos \theta, b \sin \theta)\)
- \(C = (a \cos \theta, -b \sin \theta)\)

### Step 3: Calculate the Base and Height of the Triangle
The base \(BC\) of the triangle can be calculated as the distance between points \(B\) and \(C\):
\[ BC = |y_B - y_C| = |b \sin \theta - (-b \sin \theta)| = 2b \sin \theta \]

The height \(AD\) from point \(A\) to line \(BC\) is the horizontal distance from point \(A\) to the y-axis, which is:
\[ AD = a - a \cos \theta = a(1 - \cos \theta) \]

### Step 4: Calculate the Area of Triangle \(ABC\)
The area \(A\) of triangle \(ABC\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Substituting the values we found:
\[ \text{Area} = \frac{1}{2} \times (2b \sin \theta) \times (a(1 - \cos \theta)) = ab \sin \theta (1 - \cos \theta) \]

### Step 5: Maximize the Area
To find the maximum area, we need to maximize the function:
\[ A(\theta) = ab \sin \theta (1 - \cos \theta) \]
Let \(t = \sin \theta\). Then, using the identity \(1 - \cos \theta = 1 - \sqrt{1 - t^2}\), we can express the area in terms of \(t\).

### Step 6: Differentiate and Find Critical Points
Differentiate \(A(t)\) with respect to \(t\) and set the derivative equal to zero to find critical points. This involves applying the product rule and simplifying the resulting equation.

### Step 7: Solve for \(t\)
After finding the critical points, substitute back to find the corresponding values of \(\theta\) and calculate the maximum area.

### Step 8: Final Calculation
Substituting the maximum values of \(\sin \theta\) and \(1 - \cos \theta\) into the area formula gives the maximum area of the triangle.

### Final Result
The maximum area of the isosceles triangle inscribed in the ellipse is:
\[ \text{Maximum Area} = \frac{3\sqrt{3}}{4} ab \]