Line Relection

I am deriving the equations for a reflection abount a line L, in the event someone is not familiar with them.

Let ${\bf (x_{0},y_{0}) }$ be a point on the line of reflection ${\bf L: y - y_{0} = m(x - x_{0}) }$ , where ${\bf m = tan\theta. }$ $$

Let ${\bf P(a,b) }$ be the point to be reflected and ${\bf P'(a',b') }$ be the image of ${\bf P(a,b) \implies }$

Line ${\bf L }$ is the perpendicular bisector ${\bf PP'. }$ $$

Let ${\bf M:(\frac{a + a'}{2}, \frac{b + b'}{2}) }$ be midpoint of ${\ PP'. }$

$$

${\bf y - y_{0} = m(x - x_{0}) \implies y - mx = y_{0} - mx_{0} }$ $$

Line ${\bf L }$ is perpendicular to ${\bf PP' \implies m_{PP'} = -\frac{1}{m} \implies my + x = mb + a }$

Solving the system of equations: $$

${\bf my + x = mb + a }$ $$

${\bf y - mx = y_{0} - mx_{0} }$

$$

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

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

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

$$

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

${\bf m = tan\theta = \frac{sin\theta}{cos\theta} }$

$$

Simplifying we obtain:

$$

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

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

Replacing ${\bf a, b, a', b' }$ by ${\bf x, y, x', y' }$ we obtain:

$$

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

$$

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

$$

${\bf f(x) = -x^2 }$ and ${\bf g(x) = \sqrt{x} \implies }$

${\bf \frac{d}{dx}(f(x))|_{x = a} = -2a }$ and ${\bf \frac{d}{dx}(g(x))|_{x = b} = \frac{1}{2 \sqrt{b}} \implies }$

${\bf -2a = \frac{1}{2 \sqrt{b}} \implies a = -\frac{1}{4 \sqrt{b}} }$

${\bf A: (a,-a^2) = (-\frac{1}{4 \sqrt{b}} , -\frac{1}{16 b}) }$ ${\bf B: (b, \sqrt{b} ) \implies }$

slope ${\bf m^* = \frac{1}{2 \sqrt{b}} = \frac{1}{4 \sqrt{b}} (\frac{16 b^\frac{3}{2} + 1}{4 b^\frac{3}{2} + 1}) \implies }$

${\bf 16 b^\frac{3}{2} + 1 = 8 b^\frac{3}{2} + 2 \implies 8 b^\frac{3}{2} = 1 \implies b = \frac{1}{4} }$

${\bf \implies }$ slope ${\bf m^* = 1 }$ and ${\bf a = -\frac{1}{2} }$

$$

Using ${\bf A: (-\frac{1}{2},-\frac{1}{4}) \implies y + \frac{1}{4} = x + \frac{1}{2} \implies }$

${\bf 4x - 4y + 1 = 0 }$

${\bf 4x - 4y + 1 = 0 }$ is tangent to ${\bf g(x)}$ at ${\bf B: (\frac{1}{4} , \frac{1}{2}) }$ and ${\bf f(x) }$ at ${\bf A: (-\frac{1}{2},-\frac{1}{4}) }$

$$

For ${\bf m = tan\theta = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6} \implies }$

${\bf x' = \frac{1}{2} * (x + \sqrt{3} y) }$

${\bf y' = \frac{1}{2} * (\sqrt{3} x - y) }$

$$

Using the reflection equations above: ${\bf R_({y = \frac{1}{\sqrt{3}} x }) }$ $$

For ${\bf 4 x' - 4 y' + 1 = 0 \implies 2(1 - \sqrt{3}) x + 2(1 + \sqrt{3}) y + 1 = 0 }$ and slope ${\bf m' = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}. }$

For ${\bf y' = -x'^2 \implies x^2 + 2\sqrt{3} xy + 3 y^2 + 2 \sqrt{3} x - 2 y = 0 }$

$$

and line ${\bf 4 x' - 4 y' + 1 = 0 }$ is tangent to ${\bf x^2 + 2\sqrt{3} xy + 3 y^2 + 2 \sqrt{3} x - 2 y = 0 }$ at

${\bf (-\frac{1}{4} - \frac{\sqrt{3}}{8}, -\frac{\sqrt{3}}{4} + \frac{1}{8}) }$

$$

For ${\bf y' = \sqrt{x'} \implies 3 x^2 - 2\sqrt{3} xy + y^2 - 2 x - 2\sqrt{3} y = 0 }$

$$

and line ${\bf 4 x' - 4 y' + 1 = 0 }$ is tangent to ${\bf 3 x^2 - 2\sqrt{3} xy + y^2 - 2 x - 2\sqrt{3} y = 0 }$ at

${\bf (\frac{1}{8} + \frac{\sqrt{3}}{4}, \frac{\sqrt{3}}{8} - \frac{1}{4}) }$

${\bf \therefore a = 2, \: b = 3 \implies a + b = 5. }$

$$

Note: I provided additional information that's not needed for the solution of the problem, but it was helpful in sketching the graph above.