EAMCET Engineering 2010 Solved Paper

The equation of parabola parallel to y-axis is
y=Ax2+Bx+C . . . (i)
The point (0,4),(1,9) and (4,5) lies on Eq. (i).
Then, ‌‌4=0+0+C⇒C=4 . . . (ii)
‌9=A+B+C‌
‌ or ‌‌9=A+B+4‌(∵C=4)
‌A+B=5‌‌‌...‌ (iii) ‌
‌ and ‌‌5=16A+4B+C‌‌(∵C=4)
‌ or ‌‌5=16A+4B+4‌
∴‌16A+4B=1‌
or 4A+B=‌

1

4

. . . (iv)
Solving Eqs. (iii) and (iv),
we get ‌‌A=‌

−19

12

,B=‌

79

12

. . . (v)
Substituting the values of A,B and C from Eqs. (ii) and (v) in Eq. (i), then equation of parabola is
y=‌

−19

12

x2+‌

79

12

x+4