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

Answer:

$a=-0.35747$

Step-by-step 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 $\frac{2}{tan(60^o)}$ farthar from the coordinate of point C; thus

$C_x=\frac{2}{tan(60^o)}-1=0.1547$

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

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

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

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