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
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) To find the coordinates of point A, we need to first find the equation of the parabola that the triangle lies on. Since the triangle is equilateral, the height of the triangle from vertex O to line AB will be √3 times the side length, which is 4√3.
Given that the equation of the parabola is y = ax^2, we can substitute the coordinate (2, 4√3) into the equation to find the value of a:
4√3 = a(2)^2
4√3 = 4a
a = √3
So, the equation of the parabola is y = √3x^2
Now, we can find the x-coordinate of point A by solving the equation of the parabola:
4√3 = √3(x)^2
x^2 = 4
x = ±2
Since point A lies on the right side of the y-axis, the x-coordinate of point A is 2. Substituting the x-coordinate into the equation of the parabola:
y = √3(2)^2
y = 12
Therefore, the coordinates of point A are (2, 12).
2) The value of a is √3, as calculated in the first part.
3) To find the area of equilateral triangle OAB, we can use the formula for the area of an equilateral triangle: S = (√3/4) * side length^2
Substitute the side length of OAB, which is 4, into the formula:
S = (√3/4) * 4^2
S = (√3/4) * 16
S = 4√3
Therefore, the area of equilateral triangle OAB is 4√3 units squared.