Question

Four points A,B,C and D lie in the order on the parabola $y=a{x}^2+bx+c$ and the coordinates of A, B and D are known A(-2,3); B(-1,1); D(2,7). On the basis of above information match the following :

Answer 1

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

Since points $A,B$ and $C$ lies on the curve
$\therefore$ $4a-2b+c=3,a-b+c=1,4a+2b+c=7$
Solving the equations we get $a=b=c=1$
Thus $y=x^2+x+1$
A) Value of $a+b+c=3$

B) Let $\alpha =\omega ,\beta ={\omega }^2$
Thus ${\alpha }^7+{\beta }^{19}=\omega +{\omega }^2=-1$
C) $(a+2)x^2+2\frac{(b+2)}{x}+c$
$3x^2+\frac{6}{x}+1$
Let $g\left(x\right)=3x^2+\frac{6}{x}$
$g^{\prime }\left(x\right)=6x-\frac{6}{x^2}$
$g^{\prime }\left(x\right)=0\Rightarrow x=1$
$L=3+6+1=10$
$L-3=7$
D) To maximize area of $\square ABCD$ we maximize area $\left(\Delta BCD\right)$
To maximize Area $\left(\Delta BCD\right)$ we have to maximize $h$ (as shown in figure) for maximum $h$
$\Rightarrow$ Slope of $BD$ $=$ Slope of tangent at $C$
$\frac{7-1}{2+1}=(2x+1)$
$x=\frac{1}{2}\Rightarrow y=\frac{1}{4}+\frac{1}{2}+1=\frac{7}{4}$

$c=(\frac{1}{2},\frac{7}{4})$