Question

Suppose that a parabola $y=a x^2+b x+c$, where $a>0$ and $(a+b+c)$ is an integer has vertex $\left(\frac{1}{4}, \frac{-9}{8}\right)$. If the minimum positive value of ' $a$ ' can be written as $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers, then find $(p+q)$.

Answer 1

The equation of the parabola can be taken as
$y=m\left(x-\frac{1}{4}\right)^2-\frac{9}{8} $ (given that minimum value is $\frac{-9}{8}$ when $x=\frac{1}{4}$ )
or $ y=m\left(x^2+\frac{1}{16}-\frac{x}{2}\right)-\frac{9}{8} $ or $ y=m x^2-\frac{m}{2} x+\frac{m}{16}-\frac{9}{8}$....(1)
Comparing equation (1) with $y = ax ^2+ bx + c$,
Hence $a = m , b =\frac{- m }{2}$ and $c =\frac{ m }{16}-\frac{9}{8}$
But $ a + b + c$ is an integer
$\frac{ m }{2}+\frac{ m }{16}-\frac{9}{8}=\frac{9 m -18}{16}$ is an integer.
If $a + b + c =0 \Rightarrow m =2$
If $a+b+c=-1 \Rightarrow m=\frac{2}{9}$
If $a+b+c=-2 \Rightarrow m=\frac{-14}{9}$ which is negative
Hence minimum positive value of coefficient of $x^2$ which is $m=\frac{2}{9} \Rightarrow p=2$ and $q=9$ Hence $(p+q)=11$.