TS EAMCET 20-July-2022 Shift 2 Solved Paper

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)