Given,E:y2P​=8x≡(−2,4)​andNow, directrix of Eq. (i) is x=−2So, point P(−2,4) lies on the directrix of parabola y2=8x.Hence, ∠QPQ′=2π​ (by the definition of director circle)and chord QQ′ is a focal chord and segment PQ subtends a right angle at the focus.Slope of PF=−1(∵PF⊥QQ′)Now, slope of QQ′=t1​+t2​2​=1∴PF=(2+2)2+(0−4)2​=32​=42​