Given equation of curve, ‌x2−4x+4y−8=0.....(i) ⇒x2−4x=−4y+8 ⇒x2−4x+4=−4y+8+4 ‌⇒‌‌(x−2)2=−4y+12 ‌⇒‌‌(x−2)2=−4(y−3) Now, let x−2=X and y−3=Y ∴‌‌X2=−4Y.......(ii) On comparing Eq. (i) with X2=−4aY, we get ‌4a=4 ⇒‌‌a=1 Vertex of Eq. (ii), X=0 and Y=0 ‌⇒‌‌x−2=0‌ and ‌y−3=0 ‌⇒‌‌x=2‌ and ‌y=3 ∴ Vertex of Eq.(i) is (2,3). Focus of Eq. (ii) X=0 and Y=−a ‌⇒‌‌x−2=0‌ and ‌y−3=−1 ‌⇒‌‌x=2‌ and ‌y=2 ∴ Focus of Eq. (i) is (2,2). One end of the latusrectum of Eq. (ii) ‌X=±2a‌ and ‌Y=−a ⇒‌‌x−2=±2‌ and ‌y−3=−1 ⇒‌‌x=0,4‌ and ‌y=2 ∴ One end of the latusrectum of Eq. (i) is (4,2) or (0,2) Equation of directrix of Eq. (ii), ‌Y‌=a ⇒‌y−3‌=1 ⇒‌y‌=4 and equation of axis of Eq. (ii) X=0 ⇒‌x−2‌=0 ⇒‌=2 ∴ Point of intersection of the axis and direction is (2,4)