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^o?

Answer 1

Answer:

a = -0.3575

Step-by-step explanation:

The points A and D lie on the x-axis, this means that they are the x-intercepts of the parabola, and therefore we can find their location.

The points A and B are located where

$y=a(x+1)(x-5)=0$

This gives

$x=-1$

$y=5$

Now given the coordinates of A, we are in position to find the coordinates of the point B. Point B must have y coordinate of y=2 (because the base of the trapezoid is at y=0), and the x coordinate of B, looking at the figure, must be x coordinate of A plus horizontal distance between A and B, i.e

$B_x=A_x+\frac{2}{tan(60)} =-1+\frac{2\sqrt{3} }{3}$

Thus the coordinates of B are:

$B=(-1+\frac{2\sqrt{3} }{3},2)$

Now this point B lies on the parabola, and therefore it must satisfy the equation  $y=a(x+1)(x-5).$

Thus

$2=a((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)$

Therefore

$a=\frac{2}{((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)}$

$\boxed{a=-0.3575}$