find the dimension of the largest triangle that can be inscribe in the ellipse 9x^2+16y^2=144. The sides of the rectangle are parallel to axes of the ellipse
To find the dimension of the largest triangle inscribed in the given ellipse, we need to determine the maximum distance between any two points on the ellipse.
Step 1: Convert the equation of the ellipse to its standard form:
Divide both sides of the equation by 144 to get:
x^2/16 + y^2/9 = 1
Step 2: The major axis is the longer dimension of the ellipse, and in this case, it is along the x-axis. The length of the major axis is given by 2a, where a represents the semi-major axis. In this case, a = 4.
Step 3: The minor axis is the shorter dimension of the ellipse, and it is along the y-axis. The length of the minor axis is given by 2b, where b represents the semi-minor axis. In this case, b = 3.
Step 4: The maximum distance between any two points on the ellipse is equal to the length of the diagonal of the rectangle formed by the major and minor axes. This diagonal will be the hypotenuse of a right triangle with sides of 2a and 2b.
Step 5: Using the Pythagorean theorem, we can find the length of the diagonal:
Diagonal^2 = (2a)^2 + (2b)^2
Diagonal^2 = (2 * 4)^2 + (2 * 3)^2
Diagonal^2 = 64 + 36
Diagonal^2 = 100
Step 6: Taking the square root of both sides, we find:
Diagonal = √100
Diagonal = 10
Therefore, the dimension of the largest triangle that can be inscribed in the given ellipse is 10.