Question

Four points $A,B,C$ and $D$ lie in that order on the parabola $y=l{x}^2+mx+n.$ If $A\equiv (-2,3),B\equiv (-1,1),D\equiv (2,7)$ has area of the quadrilateral $ABCD$ is maximum, then which of the following is/are true

• A. $l^2+m^2+n^2=10$
• B. $l^2+m^2+n^2=3$
• C. $C\equiv (\frac{1}{2},\frac{7}{4})$
• D. $C\equiv (1,3)$

Answer 1

Answer: (B), (C)

$C\equiv (\frac{1}{2},\frac{7}{4})$

Equation of parabola is $y=l{x}^2+mx+n$
$A,B$ and $D$ lie on the above parabola, we have
$4l-2m+n=3\cdots \left(i\right)$
$l-m+n=1\cdots \left(ii\right)$
$4l+2m+n=7\cdots \left(iii\right)$
Solving $\left(i\right),\left(ii\right),\left(iii\right)$, we get $l=m=n=1$
So, equation of parabola is $y=x^2+x+1$
Let $C\equiv (\alpha ,\beta )$
$\because A,B,D$ are fixed points, the area of quadrilateral $ABCD$ will be maximum if area of $△BCD$ is maximum.
Now,
$P($area of $△BCD)=\frac{1}{2}{\left|\begin{array}{ccc}-1& 1& 1\\ \alpha & \beta & 1\\ 2& 7& 1\end{array}\right|}_{}$
$=\frac{3}{2}(2\alpha -\beta +3)$
$P=\frac{3}{2}(-{\alpha }^2+\alpha +2)\{\because \beta ={\alpha }^2+\alpha +1\}$
$\therefore P^{\prime }=\frac{3}{2}(-2\alpha +1)$ and $P^{\prime \prime }=-3<0$
So, $P$ is maximum when $P^{\prime }=0$
$\Rightarrow \alpha =\frac{1}{2},\beta =\frac{7}{4}$
$\therefore C\equiv (\frac{1}{2},\frac{7}{4})$