Question

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

Answer 1

Given : $P:y=a{x}^2$
Let $T=(\frac{t}{2a},\frac{t^2}{4a})$
Now,
$\frac{dy}{dx}{|}_T=t=m$
So, the equation of $L$ becomes
$y=tx-b$
$T$ lies on the line, so
$\frac{t^2}{4a}=\frac{t^2}{2a}-b\Rightarrow b=\frac{t^2}{4a}\Rightarrow y=tx-\frac{t^2}{4a}\Rightarrow Q=(0,-b)=(0,-\frac{t^2}{4a})$


Now,
$TQ=1\Rightarrow \frac{t^2}{4a^2}+\frac{t^4}{4a^2}=1\Rightarrow t^2(t^2+1)=4a^2\cdots \left(1\right)$

Therefore,
$A=\frac{1}{2}\times (\frac{t^2}{4a}+\frac{t^2}{4a})\times \frac{t}{2a}-\frac{1}{\sqrt{a}}\underset{0}{\overset{\frac{t^2}{4a}}{\int }}\sqrt{y}dy\Rightarrow A=\frac{t^3}{8a^2}-\frac{2}{3\sqrt{a}}{\left(\frac{t^2}{4a}\right)}^{3/2}\Rightarrow A=\frac{t^3}{24a^2}$
Using equation $\left(1\right)$, we get
$\Rightarrow A=\frac{t^3}{6t^2(t^2+1)}\Rightarrow A=\frac{1}{6(t+\frac{1}{t})}$
So,
$M=\frac{1}{6\times 2}\therefore \frac{1}{M}=12$