1)
Since triangle OAB is equilateral, we know that angle OAB is 60 degrees. This means that the coordinate of point A is at the solution of the equation y = ax^2 where the tangent of the angle between the parabola and the x-axis is √3.
At point A, the slope of the tangent line to the parabola is equal to the derivative of y = ax^2 at point A.
The derivative of y = ax^2 is 2ax.
At point A, y = a(x)^2 and the tangent line has slope 2ax. Therefore, 2ax = √3 and we know that x = 2 so a = √3/4.
2) Substituting the coordinates of point A (x = 2, y = 3) into y = ax^2:
3 = (√3/4)*(2)^2
3 = (√3/4)*4
3 = 3√3
a = √3/4
3) To find the area of the equilateral triangle OAB, we can use the formula:
Area = (side length)^2 * (√3/4)
Area = 4^2 * (√3/4)
Area = 16 * √3
Therefore, the area of the equilateral triangle OAB is 16√3.
When equilateral triangle OAB with sides of length 4 lies on parabola y=ax squared(a>0) as shown below, solve the following problems.
1) find the coordinates of point A
OC=
Since AC:4= squared root of 3:2
Ac=
Therefore A ( ,)
2) Find the value of a
substituting the coordinates of point A into y=ax squared
3) Find the area S of equilateral triangle OAB
1 answer