A rectangle with sides parallel to the coordinate axes is inscribed in
the ellipse
x2/a2 + y2/b2 = 1:
Find the largest possible area for this rectangle.
To find the largest possible area for the rectangle inscribed in the ellipse, we can approach the problem using calculus.
Let's assume the sides of the rectangle are parallel to the coordinate axes, so the rectangle's sides are horizontal and vertical. We can represent the rectangle's sides as x and y, respectively.
The area of the rectangle is given by A = x * y. To maximize the area, we need to find the values of x and y that satisfy the equation of the ellipse while maximizing the area.
The equation of the ellipse is given as x^2/a^2 + y^2/b^2 = 1, where a and b are the lengths of the ellipse's major and minor axes, respectively.
We need to rewrite the equation of the ellipse in terms of x and y, where x and y represent the coordinates of the rectangle's vertices.
Dividing both sides of the equation by a^2, we get:
x^2/a^2 + y^2/b^2 = 1
Rearranging the equation, we get:
x^2/a^2 = 1 - y^2/b^2
Multiplying both sides of the equation by a^2, we get:
x^2 = a^2 - (a^2/b^2) * y^2
Now, we have the equation of the ellipse in terms of x and y, and we can use it to express x in terms of y and find the values of x and y that maximize the area.
For a rectangle, the opposite vertices will have coordinates (±x/2, ±y/2), where x and y are the sides of the rectangle. We want to find the maximum values of x and y that satisfy the equation of the ellipse.
Substituting x^2 = a^2 - (a^2/b^2) * y^2 into the area equation, A = x * y, we get:
A = (a^2 - (a^2/b^2) * y^2) * y
To maximize A, we can take the derivative of A with respect to y, set it to zero, and solve for y.
dA/dy = 0
Differentiating A with respect to y, we get:
dA/dy = (a^2 - (a^2/b^2) * y^2) + (-2 * (a^2/b^2) * y) = 0
Simplifying the equation, we would get:
a^2 - (a^2/b^2) * y^2 - (2 * a^2/b^2) * y = 0
Now, we can solve this quadratic equation for y to find the y-coordinate that maximizes the area.
Using the quadratic formula, we can solve for y:
y = (-(-2 * a^2/b^2) ± sqrt((-2 * a^2/b^2)^2 - 4 * (a^2/b^2) * (a^2 - A)))/(2 * (a^2/b^2))
Simplifying the equation even further, we get:
y = (2 * a^2 ± sqrt(4 * a^4 - 4 * (a^2/b^2) * (a^2 - A)))/(2 * (a^2/b^2))
y = (a^2 ± sqrt(a^2(a^2/b^2 - 4 + 4A/b^2)))/(a^2/b^2)
y = (± sqrt(a^2/b^2 * (a^2 - 4 + 4A/b^2)))/(a^2/b^2)
y = ± sqrt(a^2 - 4 + 4A/b^2)
Since the rectangle lies in the first quadrant, we consider the positive square root:
y = sqrt(a^2 - 4 + 4A/b^2)
Now that we have obtained y, we can substitute this value into the equation of the ellipse to find x:
x = sqrt(a^2 - (a^2/b^2) * (a^2 - 4 + 4A/b^2))
The area of the rectangle can be calculated using the formula A = x * y.
By calculating x and y using the equations above and then substituting these values into A = x * y, we can determine the largest possible area for the rectangle inscribed in the ellipse.