Question

Let $L$ be the line obtained by rotating the tangent line, drawn to the parabola $y=x^2$ at the point $A(1,1)$ about the point $A$ by an angle of ${45}^{\circ }$ in the clockwise direction. Let $B$ bet the intersection of the line $L$ with $y=x^2$ other than $A$. If the area enclosed by the line $L$ and the parabola is $K$ sq.unit, then $\left[K\right]$ is equal to, $($ where $[\cdot ]$ denotes greatest integer function $)$:

Answer 1

Parabola: $y=x^2\cdots \left(i\right)$
$(a=\frac{1}{4})$
Tangent at $A(1,1)$ is $2x=y+1\cdots \left(ii\right)$.
Slope of the tangent at $A(1,1)=2$
angle between $L$ and the tangent is ${45}^{\circ }$, and let slope of $L$ be $m$
So, $\tan \left(\frac{\pi }{4}\right)=\frac{2-m}{1+2m}$
$\Rightarrow m=\frac{1}{3}$
$L:y=\frac{1}{3}x+c$
$L$ passes through $A(1,1)$
$\Rightarrow 1=\frac{1}{3}\left(1\right)+c\Rightarrow c=\frac{2}{3}$
$B$ is point of intersection other than $A$
$\Rightarrow B=(\frac{-2}{3},\frac{4}{9})$

Required area $=\underset{-2/3}{\overset{1}{\int }}\left(\frac{x+2}{3}\right)dx-\underset{-2/3}{\overset{1}{\int }}{x}^{2}dx$
$=\frac{1}{3}{[\frac{x^2}{2}+2x]}_{-2/3}^1-{\left[\frac{x^3}{3}\right]}_{-2/3}^1=\frac{1}{3}[\frac{1}{2}-\frac{4}{18}+2(1)+\frac{4}{3}]-[\frac{1}{3}+\frac{8}{81}]=\frac{125}{2(3{)}^4}=\frac{125}{162}=K$
$\therefore \left[K\right]=0$