Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis.

To find the maximum area of an isosceles triangle inscribed in an ellipse with its vertex at one end of the major axis, we start by defining the ellipse's equation. For an ellipse centered at the origin, the equation is given by:

Ellipse Equation

The standard form is:

(x²/a²) + (y²/b²) = 1

Here, a is the semi-major axis and b is the semi-minor axis.

Triangle Configuration

Assume the vertex of the triangle is at point (a, 0) on the major axis. The other two vertices will be symmetric about the y-axis, located at points (-x, y) and (x, y).

Area Calculation

The area A of the triangle can be calculated using the formula:

A = (1/2) * base * height

In this case, the base is 2x and the height is y, so:

A = x * y

Using the Ellipse Equation

From the ellipse equation, we can express y in terms of x:

y = b * sqrt(1 - (x²/a²))

Substituting this into the area formula gives:

A = x * b * sqrt(1 - (x²/a²))

Maximizing the Area

To find the maximum area, we differentiate A with respect to x and set the derivative to zero:

dA/dx = 0

Solving this will yield the optimal value of x. After finding x, substitute it back to find y and then calculate the maximum area.

Final Result

The maximum area of the isosceles triangle inscribed in the ellipse is:

A_max = (b * a) / 2

This area occurs when the triangle's base is aligned with the ellipse's width, maximizing the height from the vertex to the base.