∴ Focus (ae,0)=(√2,0) Equation of tangent, y=mx+√a2m2+b2 y=mx+√4m2+2 Passes through (h, k) (k−mh)2=4m2+2 ....(1) line perpendicular to tangent will have slope −
1
m
y−0=−
1
m
(x−√2) my=−x+√2 (h+mk)2=2.....(2) Add equation (1) and (2) k2(1+m2)+h2(1+m2)=4(1+m2) h2+k2=4 x2+y2=4 (Auxiliary circle) ∴(−1,√3) lies on the locus.