Let, B≡(p,0)

⇒AB=p+1,p>1
Now, finding point C using parametric form we get,
cos32πx+1=sin32πy−0=p+1
⇒C≡(2−p−23,23(p+1))
⇒BC2=(2−3(p+1))2+(23(p+1))2
⇒BC=3(p+1)=43
⇒p=3
So, C≡(−3,23)
Now, equation of line BC is given by,
y=3−1(x−3)
Solving with line y=x+3.
⇒x+3=33−x
⇒x3+33=3−x
⇒x(3+1)=3(1−3)
⇒x=1+33(1−3)
⇒x=2−3(4−23)
⇒α=−3(2−3)andβ=−3+33
⇒α2β4=9(2−3)281(3−1)4
⇒α2β4=(2−3)29(4−23)2
⇒α2β4=36