Question

The parabola $P:y=a{x}^2,$ where $a$ is a positive real constant, is touched by the line $L:y=mx-b$ (where $m$ is a positive constant and $b$ is real) at the point $T.$ Let $Q$ be the point of intersection of the line $L$ and $y-$axis such that $TQ=1.$ If $A$ denotes the maximum value of the region surrounded by $P,$ $L$ and the $x-$axis, then the value of $\frac{1}{A}$ is

Answer 1

$y=a{x}^2$
$\frac{dy}{dx}=2ax$
At $T(x_0,a{x}_0^2),$ we have $\frac{dy}{dx}{|}_{x=x_0}=2a{x}_0$
Thus, the equation of tangent is
$y-a{x}_0^2=2a{x}_0(x-x_0)$
$\Rightarrow y=2ax{x}_0-a{x}_0^2$
$m=2a{x}_0$ and $b=a{x}_0^2$
$\therefore Q(0,-a{x}_0^2)$
Given : $TQ=1$
$\Rightarrow \sqrt{(x_0-0{)}^2+(a{x}_0^2+a{x}_0^2{)}^2}=1$
$\Rightarrow x_0^2+4a^2{x}_0^4=1$
$\Rightarrow a^2=\frac{(1-x_0^2)}{4x_0^4}\cdots \left(1\right)$

Now, bounded area is
$I=\underset{0}{\overset{x_0}{\int }}(a{x}^2-mx+b)dx$
$\Rightarrow I={|\frac{a{x}^3}{3}-\frac{m{x}^2}{2}+bx|}_0^{x_0}$
$\Rightarrow I=\frac{a{x}_0^3}{3}-\frac{m{x}_0^2}{2}+b{x}_0$
$\Rightarrow I=\frac{a{x}_0^3}{3}-\frac{2a{x}_0^3}{2}+a{x}_0^3=\frac{a{x}_0^3}{3}$
$\Rightarrow I=\frac{x_0\sqrt{1-x_0^2}}{6}$
Differentiating w.r.t. $x_0,$ we get
$\frac{dI}{d{x}_0}=\frac{1}{6}\{\sqrt{1-x_0^2}+\frac{1}{2}\frac{-2x_0^2}{\sqrt{1-x_0^2}}\}$
$\Rightarrow \frac{dI}{d{x}_0}=\frac{1}{6}\left(\frac{1-2x_0^2}{\sqrt{1-x_0^2}}\right)$
For max/min, $\frac{dI}{d{x}_0}=0$
$\Rightarrow 1-2x_0^2=0$
$\Rightarrow x_0=\pm \frac{1}{\sqrt{2}}$
Sign scheme of $\frac{dI}{d{x}_0}:$

$\therefore$ Area is maximum at $x_0=\frac{1}{\sqrt{2}}$
$I_{max}=\frac{1}{12}=A$
Hence, the value of $\frac{1}{A}$ is $12.$