Question

Consider the collection of all curves of the form $y=a-b{x}^2$ that pass through the point $(2,1)$ where $a$ and $b$ are positive real numbers. If the minimum area of the region bounded by $y=a-b{x}^2$ and the $x-$axis is $\sqrt{A}$ sq. units then the value of $A\in N$ is

Answer 1

$y=a-b{x}^2$
When $x=2,y=1$
$\Rightarrow 1=a-4b\Rightarrow a=1+4b$

Now, Area $A=2\underset{0}{\overset{\sqrt{a/b}}{\int }}(a-b{x}^2)dx$
$\Rightarrow A=2{[ax-\frac{b{x}^3}{3}]}_0^{\sqrt{a/b}}$
$\Rightarrow A=2[\frac{a\sqrt{a}}{\sqrt{b}}-\frac{b}{3}\cdot \frac{a\sqrt{a}}{b\sqrt{b}}]$
$\Rightarrow A=\frac{4a^{3/2}}{3\sqrt{b}}$
$\Rightarrow A=\frac{4}{3}\cdot \frac{(1+4b{)}^{3/2}}{\sqrt{b}}(\because a=1+4b)$
Differentiating w.r.t. $b,$ we get
$\frac{dA}{db}=\frac{4}{3}\left[\frac{6\cdot b^{1/2}\cdot (1+4b{)}^{1/2}-(1+4b{)}^{3/2}\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{b}}}{b}\right]$
$\Rightarrow \frac{dA}{db}=\frac{2}{3}\left[\frac{12b\sqrt{(1+4b)}-(1+4b{)}^{3/2}}{b^{3/2}}\right]$
$\Rightarrow \frac{dA}{db}=\frac{2}{3}\frac{\sqrt{1+4b}}{b^{3/2}}(12b-1-4b)$

For max/min area, $\frac{dA}{db}=0$
$\Rightarrow b=\frac{1}{8};a=\frac{3}{2}$
By first derivative test, $A$ has $b=\frac{1}{8}$ as point of minima.
$\therefore A_{min}=\frac{4}{3}\cdot {\left(\frac{3}{2}\right)}^{3/2}\cdot 2\sqrt{2}$
$\Rightarrow A_{min}=\sqrt{48}$ sq. units