So, Focus (F)=(±ae,0)=(±√6,0) Now, equation of tangent at P(4,√6) is 4x−2√6y=4[∵x⋅x−2⋅y⋅y=4,x⋅4−2⋅y⋅√6=4] ⇒2x−√6y=2⋅⋅⋅⋅⋅⋅⋅(i) Putting y=0 in Eq. (i), we get x -intercept of tangent i.e. x=1 ∴Q≡(1,0) Hence, equation of corresponding latus rectum is x=√6 . ∴R≡(√6,