The correct option is B −2
Solving the equations x2=ay and y−2x=1, we get
x2=a(2x+1)
⇒x2−2ax−a=0
So, the given line cuts the parabola at two points (x1,y1) and (x2,y2)
∴x1+x2=2a and x1x2=−a
Now, (√40)2=(x1−x2)2+(y1−y2)2⇒40=(x1−x2)2+(x21a−x22a)2⇒40=(x1−x2)2[1+(x1+x2)2a2]⇒40=[(x1+x2)2−4x1x2](4a2a2+1)⇒40=5(4a2+4a)⇒a2+a−2=0
⇒(a+2)(a−1)=0∴a=1,−2