Suppose that a parabola y=a x2+b x+c, where a0 and (a+b+c) is an integer has vertex ((1/4), (-9/8)). If the minimum positive value of ' a ' can be written as (p/q) where p and q are relatively prime positive integers, then find (p+q).

Question

Suppose that a parabola y=a x2+b x+c, where a>0 and (a+b+c) is an integer has vertex ((1/4), (-9/8)). If the minimum positive value of ' a ' can be written as (p/q) where p and q are relatively prime positive integers, then find (p+q). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The equation of the parabola can be taken as

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

(given that minimum value is

\frac{- 9}{8}

when

x = \frac{1}{4}

)
or

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

or

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

....(1)
Comparing equation (1) with

y = a x^{2} + b x + 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

.

Q. Suppose that a parabola y=ax2+bx+c, where a>0 and (a+b+c) is an integer has vertex (41​,8−9​). If the minimum positive value of ' a ' can be written as qp​ where p and q are relatively prime positive integers, then find (p+q).

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