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Suppose that a parabola has vertex $(\frac{1}{4}, \frac{- 9}{8})$ and the equation $y = a x^{2} + b x + c,$ where a > 0 and a + b + c is an integer. The minimum possible value of a can be written in the form $\frac{p}{q}, (G C D (p, q) = 1)$ then p + q is equal to

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answer is C.

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Detailed Solution

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

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

Comparing with $y = a x^{2} + b x + c$

$a = a, b = \frac{- a}{2}, c = \frac{a}{16} - \frac{9}{8}$

$a + b + c = a + (\frac{- a}{2}) + (\frac{a}{16} - \frac{9}{8}) = \frac{9 a}{16} - \frac{9}{8}$

a + b + c is an integer $\Rightarrow 9 a - 18$ is divisible by 16

$9 a = 18 + 16 λ$

$λ = - 1 \Rightarrow 9 a = 2$

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

$p + q = 11$

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