Question

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

Answer 1

Here, the equation of the parabola is,

$y=a{x}^2+bx+c$, where $a>0$ and $a+b+c$ is an integer

Vertex is at $(\frac{1}{4},-\frac{9}{8})$

The equation of the parabola can be expressed as,

$y=a{(x-\frac{1}{4})}^2-\frac{9}{8}$

Expand the above equation.

$y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}$

$y=a{x}^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}$

From the above result and the equation given in the question, we have

$a=a$

$b=-\frac{a}{2}$

$c=\frac{a}{16}-\frac{9}{8}$

Add all the above results.

$a+b+c=\frac{9a-18}{16}$

Since, $a+b+c$ is an integer, $9a-18$ must be divisible by $16$.

Let $9a=z$. Now, if

$z-18\equiv 0(mod16)$

Then,

$z\equiv 2(mod16)$

Therefore, if

$9a=2$

$a=\frac{2}{9}=\frac{p}{q}$

Therefore,

$p+q=2+9=11$

Hence, this is the required result.