Question

The graph of $y=a{x}^2+bx+c$ is shown in the figure below, where $Q$ is the vertex of the parabola. If the length of $PQ$ and $OR$ are $9\text{units}$ and $5\text{units}$ respectively and area of $△OBQ$ is $\frac{45}{4}\text{sq. units}$, then which of the following is/are correct?

• A. length of $AB$ is $3\text{units}$
• B. $\left|\frac{b}{a}\right|<1$
• C. length of $AB$ is $4\text{units}$
• D. $\left|\frac{b}{a}\right|>1$

Answer 1

Answer: (A), (D)

$\left|\frac{b}{a}\right|>1$

Given : $y=a{x}^2+bx+c$


From the graph
$OR=5\Rightarrow c=-5PQ=9\Rightarrow -\frac{D}{4a}=-9\Rightarrow \frac{b^2+20a}{4a}=9\Rightarrow b^2=16a\cdots \left(1\right)$
Now,
$\text{Area of}△OQB=\frac{45}{4}\Rightarrow \frac{1}{2}\times OB\times 9=\frac{45}{4}\Rightarrow OB=\frac{5}{2}$
Therefore, one root of the quadratic equation is $\frac{5}{2}$, so
$a{\left(\frac{5}{2}\right)}^2+b\left(\frac{5}{2}\right)-5=0\Rightarrow 5a+2b-4=0\cdots \left(2\right)$
Using equation $\left(1\right)$, we get
$\Rightarrow 5\left(\frac{b^2}{16}\right)+2b-4=0\Rightarrow 5b^2+32b-64=0\Rightarrow 5b^2+40b-8b-64=0\Rightarrow (b+8)(5b-8)=0\Rightarrow b=-8,\frac{8}{5}$
Using equation $\left(2\right)$, we get
$\Rightarrow a=4,\frac{4}{25}$
As $-\frac{b}{2a}>0$, so
$b=-8,a=4\Rightarrow \left|\frac{b}{a}\right|=2>1\Rightarrow y=4x^2-8x-5\Rightarrow y=(2x+1)(2x-5)\Rightarrow y=0\Rightarrow x=-\frac{1}{2},\frac{5}{2}\therefore AB=3\text{units}$