Consider a parabola y=x24 and the point F(0,1). Let A1(x1,y1),A2(x2,y2),A3(x3,y3),...,Ak(xk,yk) are 'n' points on parabola such as xk>0 and ∠OFAk=kπ2n,(k=1,2,3,...,n). Then the value of limn→∞1nn∑k=1FAk is
A
2π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4π
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
16π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B4π Let Ak=(2t,t2)
Slope of FAk=t2−12t−0=tan(π2+θk) ⇒tanθk=2t1−t2=tan(2ϕ) (Say) ⇒ϕ=θk2=kπ4n where tanϕ=t Also FAk=√(t2−1)2+(2t)2⇒FAk=t2+1⇒FAk=1+tan2ϕ⇒=sec2(kπ4n) ∴limn→∞1nn∑k=1FAk=limn→∞1nn∑k=1sec2(π4(kn)) =1∫0sec2(πx4)dx=4π[tanπx4]10=4π