Question

An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if points A and D belong to the x−axis and m∠BAD=60 degrees?

Answer 1

Answer:

  $a=-\dfrac{3+9\sqrt{3}}{52}\approx −0.357470332079$

Step-by-step explanation:

Here, we assume that point A is (-1, 0) and point B is (z, 2). Since angle BAD is 60°, we have ...

  2/(z+1) = tan(60°) = √3

  z = (2/√3) -1

Putting the value of z into the equation for the parabola lets us find "a".

  2 = a(z +1)(z -5) = a(2/√3)(2/√3 -6)

Then ...

  a = 2/(4/3 -12/√3) = (6√3)/(4√3 -36)

  a = -(3+9√3)/52 ≈ −0.357470332079