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

The value of a is 0.

To find the value of a, let's analyze the given information about the isosceles trapezoid ABCD.

Since points A and D belong to the x-axis, their y-coordinates are both zero. The parabola equation

y=a(x+1)(x−5) can be used to find the x-coordinates of points A and D.

Let's first find the x-coordinate of point A:

0=a(0+1)(0−5)

0=−6a

a=0

Now, let's find the x-coordinate of point D:

0=a (d+1) (d−5)

where d is the x-coordinate of point D.

Given that the height of the trapezoid is 2 units, and m∠BAD is 60 degrees, we can use trigonometry to find the length of the trapezoid's bases.

The triangle ABD is an equilateral triangle (since m∠BAD = 60 degrees), so the bases of the trapezoid are both 2 units.

Now, we can set up the equation:

0=a (2+1 )(2−5)

0=−3a

So,

a=0, or a ≠0, but we know a ≠0 from the initial parabola equation.

Answer 2

`a=-(3+9√(3))/(52)\approx −0.357470332079`

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