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=60o?

Answer 1

`a=-0.35747`

Explanation:

We know that all the vertices of the isosceles trapezoid lie on the parabola, and the points A and D lie along the x-axis, i.e at
`y=0`

Therefore points A and D are located where

`a(x+1)(x-5)=0`

`A=x=-1`

`D=x=5`

Now we need to find the coordinates of point C; we already have its y-coordinate (it's
`y=2`), and looking at the figure attached we see that the x-coordinate of point C is
`(2)/(tan(60^o))` farthar from the coordinate of point C; thus

`C_x=(2)/(tan(60^o))-1=0.1547`

Therefore the coordinates of C are
`C=(0.1547,2)`

Now this point C lies on the parabola, and therefore must satisfy the equation
`y=a(x+1)(x-5):`

`2=a(0.1547+1)(0.5147-5)`

`\therefore a=(2)/((0.1547+1)(0.5147-5)) =-0.35747\\\\\boxed{a=-0.35747}`