The length of the chord intercepted from line y=mx+c by parabola x2=4ay is 4√a(1+m2)(c+am2) So √40‌‌=4√a(1+4)(1+a(4)) 40‌‌=16(5a(1+4a)) 1‌‌=2a(1+4a) 8a2+2a−1‌‌=0 Further simplify the above, 8a2+4a−2a−1‌‌=0 4a(2a+1)−1(2a+1)‌‌=0 a‌‌=‌