Dilations and Reflections!

Prior to doing the I will quickly derive the equations of a reflection of a point about some line.

The given line has slope $m \implies m_{PP'} = m_{\perp} =-\dfrac{1}{m} \implies$

$my + x = bm + a$

and the given line is:

$y - mx = -mx_{0} + y_{0}$

Solving the system we obtain:

$x = \dfrac{m^2x_{0} + (b - y_{0})m + a}{m^2 + 1} = \dfrac{a + a'}{2}$

$y = \dfrac{bm^2 + (a - x_{0})m + y_{0}}{m^2 + 1} = \dfrac{b + b'}{2}$

Solving for $a'$ and $b'$ we obtain:

$a' = \dfrac{2m^2x_{0} + 2(b - y_{0})m + (1 - m^2)a}{m^2+ 1}$

$b' = \dfrac{bm^2 + 2(a - x_{0})m + 2y_{0} - b}{m^2 + 1}$

$m = \tan(\theta) \implies$

$a' = (a - x_{0})\cos(2\theta) + (b - y_{0})\sin(2\theta) + x_{0}$

and

$b' = (a - x_{0})\sin(2\theta) - (b - y_{0})\cos(2\theta) + y_{0}$

Replacing $a,b,a'$ and $b'$ by $x,y,x'$ and $y'$ respectively we obtain:

$x' = (x - x_{0})\cos(2\theta) + (y - y_{0})\sin(2\theta) + x_{0}$

$y' = (x - x_{0})\sin(2\theta) - (y - y_{0})\cos(2\theta) + y_{0}$

$m = \tan(\theta) = \sqrt{3} \implies \theta = \dfrac{\pi}{3}$ and since $y = \sqrt{3}x$ passes thru the origin, using the equations of reflection above with a dilation factor of $\dfrac{1}{2}$ , we obtain:

$x = \dfrac{1}{2}(-\dfrac{1}{2}x' + \dfrac{\sqrt{3}}{2}y')$

$y = \dfrac{1}{2}(\dfrac{\sqrt{3}}{2}x' + \dfrac{1}{2}y')$

Using the above equations for the parabola $x^2 - 2\sqrt{3}xy + 3y^2 - \sqrt{3}x - y = 0$ we obtain:

$\dfrac{1}{4}x'^2 - \dfrac{\sqrt{3}}{2}x'y' + \dfrac{3}{4}y'^2$

$+ \dfrac{3}{2}x'^2 - \sqrt{3}x'y' - \dfrac{3}{2}y'^2$

$+ \dfrac{9}{4}x'^2 + \dfrac{3\sqrt{3}}{2}x'y' + \dfrac{3}{4}y'^2$

$+ \sqrt{3}x' - 3y' - \sqrt{3}x' - y' = 0$

$\implies 4x'^2 - 4y' = 0 \implies \boxed{y' = x'^2}$

and the center of the given circle $(\dfrac{\sqrt{3}}{4}a,\dfrac{a}{4})$ $\rightarrow (0,a)$ in the $x'y'$ plane.

Using $y' = x'^2$ and center $O':(0,a)$ and $P':(x',y') = (x',x'^2) \implies$

$D = r^2 = x'^2 + (x'^2 - a)^2 \implies \dfrac{dD}{dx} = 2x'(2x'^2 + 1 - 2a) = 0$

$x' \neq 0 \implies x' = \pm\sqrt{\dfrac{2a - 1}{2}}$ and the radius $r = 1 \implies 4a - 1 = 4 \implies$

$a = \dfrac{5}{4} \implies x' = \dfrac{\sqrt{3}}{2} \implies y' = \dfrac{3}{4} \implies$ the equation of the circle is

$x'^2 + (y' - \dfrac{5}{4})^2 = 1$ and the portion of the circle needed is $\boxed{y' = \dfrac{5}{4} - \sqrt{1 - x'^2}}$ .

The area $A = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} (\dfrac{5}{4} - \sqrt{1 - x'^2} - x'^2) dx$

Letting $x' = \sin(\lambda) \implies dx' = \cos(\lambda) d\lambda \implies$

$A = 2(-\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi}{3}} (1 + \cos(2\lambda)) d\lambda + (\dfrac{5}{4}x' - \dfrac{x'^3}{3})|_{0}^{\frac{\sqrt{3}}{2}}) =$

$2(-\dfrac{1}{2}(\lambda + \dfrac{1}{2}\sin(2\lambda))|_{0}^{\frac{\pi}{3}} + \dfrac{\sqrt{3}}{2})$

$= \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3} = \dfrac{a\sqrt{a}}{b} - \dfrac{\pi}{a} \implies a + b = \boxed{7}$ .