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 ${45}^o$ 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 $\frac{l}{m}$ (where $l,m$ are integers and are coprime) then the unit digit in $l+m$ is:

Answer 1

Parabola: $y=x^2..........\left(1\right)(a=\frac{1}{4})$

Tangent at $A(1,1)$ is $2x=y+1...............\left(2\right)$

Slope of equation $\left(2\right)=2$

angle between $L$ and equation $\left(2\right)$ is $45°$, and slope of $L$ be$=m$

So, $tan\left(\frac{\pi }{4}\right)=\left|\frac{2-m}{1+2m}\right|$

$=>m=\frac{1}{3}$

$L:y=\frac{1}{3}x+c$

$L$ passes through $A(1,1)$

$=>-1=\frac{1}{3}\left(1\right)+c$

$=>c=\frac{2}{3}$

$B$ is point of intersection other than $A=>B:(\frac{-2}{3},\frac{4}{9})$

Area $BOA={\int }_{\frac{-2}{3}}^1\left(\frac{x+2}{3}\right)dx-{\int }_{\frac{-2}{3}}^1{x}^{2}dx$

$=>\frac{l}{m}=\frac{1}{3}[\frac{x^2}{2}+2x{]}^1-[\frac{x^3}{3}{]}^1$

$=>\frac{l}{m}=\frac{1}{3}[\frac{1}{2}-\frac{4}{18}+2(1)+\frac{4}{3}]-[\frac{1}{3}+\frac{8}{18}]$

$=>\frac{l}{m}=\frac{125}{2(3{)}^4}=\frac{125}{162}$

$l=125,m=162$

$=>l+m=125+162=287$