AP EAPCET 04-Jul-2022 Shift 2 Paper

(a) Let the vertex of the parabola be the point (h,k) and length of its latus rectum be 4a. Since, its axis is parallel to Y - axis, its equation can be written as
(x−h)2=4a(y−k)......(i)
It passes through the given points (0,4),(1,9) and (4,5).
∴‌‌(0−h)2‌=4a(4−k)
⇒‌‌h2‌=4a(4−k).....(ii)
(1−h)2‌=4a(9−k)
⇒1−2h+h2‌=4a(9−k).......(iii)
‌ and ‌‌‌(4−h)2‌=4a(5−k)
16−8h+h2‌=4a(5−k)......(iv)
Now, subtracting Eqs. (ii) and (iii), Eqs. (iii) and (iv), we get
1−2h=20a......(v)
and ‌‌15−6h=−16a......(vi)
(vi)
Solving Eqs. (v) and (vi), we get a=−‌

3

19

,h=‌

79

38

Putting these value to get k=‌

9889

912

Thus, equation of parabola is
‌(x−‌

79

38

)2=‌

−12

19

(y−‌

9889

912

)
⇒19x2+12y−79x−48=0