Find the equation to the parabola, whose axis parallel to the $y$-axis and which passes through the points…
Find the equation to the parabola, whose axis parallel to the $y$-axis and which passes through the points $(0,4),(1,9)$ and $(4,5)$ is
$y=-x^2+x+4$
$y=-x^2+x+1$
$y=\frac{-19}{12} x^2+\frac{79}{12} x+4$
$y=\frac{-19}{12} x^2+\frac{89}{12}+1$
Solution
The equation of parabola parallel to $y$-axis is
$y=A x^2+B x+C$ ...(i)
The point $(0,4),(1,9)$ and $(4,5)$ lies on Eq. (i). Then, $\quad 4=0+0+C \Rightarrow C=4$ ...(ii)
$9=A+B+C$
or $\quad 9=A+B+4 \quad(\because C=4)$
$A+B=5$ ...(iii)
and $\quad 5=16 A+4 B+C \quad(\because C=4)$
or $\quad 5=16 A+4 B+4$
$\therefore \quad 16 A+4 B=1$
$4 A+B=\frac{1}{4}$ ...(iv)
Solving Eqs. (iii) and (iv),
we get $\quad A=\frac{-19}{12}, B=\frac{79}{12}$ ...(v)
Substituting the values of $A, B$ and $C$ from Eqs. (ii) and (v) in Eq. (i), then equation of parabola is
$y=\frac{-19}{12} x^2+\frac{79}{12} x+4$