‌C:x2+2x−4y+9=0 ‌C:(x+1)2=4(y−2) ‌TP(1,3):x⋅1+(x+1)−2(y+3)+9=0 ‌:2x−2y+4=0 ‌Tp:x−y+2=0 ‌A:(0,2) ‌‌ Line ‌|‌ to ‌x−3y=6‌ passes ‌(1,3)‌ is ‌x−3y+8=0 ‌‌ Meet parabola ‌:y2=4x ‌⇒y2=4(3y−8) ‌⇒y2−12y+32=0 ‌⇒(y−8)(y−4)=0 ‌⇒‌ point of intersection are ‌ ‌(4,4)&(16,8)‌ lies on ‌2x−3y=8 ‌‌ Hence ‌A:(0,2) ‌‌ B : ‌(16,8) ‌(AB)2=256+36=292