The correct option is D Length of the rectangle is 5√58
Let the coordinates of P(p,p2) and R(r,r2)
∴S=(p+r2,pr)
Slope of the tangent,
y=x2⇒y′=2x
We know that the line SP is perpendicular to SR, so
2p×2r=−1⇒r=−14p ⋯(1)
It is given that,
SP=2SR⇒SP2=4SR2
⇒(p−r2)2+(p2−pr)2=4((r−p2)2+(r2−pr)2)
⇒−34(p−r)2=(p−r)2[4r2−p2]⇒p2−4r2=34[∵p≠r]
Using equation (1),
⇒p2−416p2=34⇒4p4−3p2−1=0⇒(p2−1)(4p2+1)=0⇒p=±1⇒r=∓14
Now, the coordinate of
P=(−1,1) or (1,1)
R=(14,116) or (−14,116)S=(−38,−14) or (38,−14)
Now, area of the rectangle PQRS
=(PS)×(SR)=(PS)22=12[(p−r2)2+(p2−pr)2]=12[(p−r)2(p2+14)]=12×5242×54=125128∴PS=√12564=5√58