Question

The points $A,B,C$ and $D$ lies on the parabola $y=a{x}^2+bx+c$, where the coordinates of points $A,B$ and $D$ is $A(-2,14),B(-1,7)$ and $D(2,10)$. The ordinate of $C$, for which the area of the quadrilateral $ABCD$ is greatest, is

Answer 1

Let the co-ordinates of $C$ be $(x,y)$
Since points $A,B$ and $D$ lie on $y=a{x}^2+bx+c$,
$14=4a-2b+c\cdots \left(i\right)$
$7=a-b+c\cdots \left(ii\right)$
$10=4a+2b+c\cdots \left(iii\right)$
Solving them,
$\Rightarrow a=2,b=-1,c=4$
$\Rightarrow y=2x^2-x+4$
Area of quadrilateral will be least when area of $\Delta BCD$ is least, so
$A=\frac{1}{2}\left|\begin{array}{ccc}-1& 7& 1\\ x& y& 1\\ 2& 10& 1\end{array}\right|$
$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=\frac{1}{2}\left|\begin{array}{ccc}-1& 7& 1\\ x+1& y-7& 0\\ 3& 3& 0\end{array}\right|$
$=\frac{1}{2}|3x+3-3y+21|$
$=\frac{1}{2}|3x-6x^2+3x-12+24|$
$=\frac{1}{2}|-6x^2+6x+12|$

$A=3|x^2-x-2|=3|(x+1)(x-2)|\Rightarrow \frac{dA}{dx}=\pm 3(2x-1)=0\Rightarrow x=\frac{1}{2}$

$A=\{\begin{array}{c}3(x+1)(x-2)x\in (-\infty ,-1)\cup (2,\infty )\\ -3(x+1)(x-2)x\in (-1,2)\end{array}$

So for $x=\frac{1}{2}$
$A=-3(x^2-x-2)\Rightarrow \frac{d^{2}A}{d{x}^2}=-6<0$

$y=2\times \frac{1}{2^2}-\frac{1}{2}+4=4$
$C\equiv (\frac{1}{2},4)$
Hence the value of the ordinate will be $4$.