Answer
See graph and explanations.
Work Step by Step
Step 1. Fermat's principle states that “light travels between two points along the path that requires the least time, as compared to other nearby paths.”
Step 2. See figure; let the mirror be the x-axis, the light source be at $A(0,a)$ and the light receiver be at $B(b,c)$ where $a,b,c$ are positive constants.
Step 3. Assume the reflection point is at $R(x,0), 0\lt x\lt b$; we can find distances $AO=\sqrt {x^2+a^2}$ and $BO=\sqrt {(b-x)^2+c^2}$. The total distance is then $s=AO+BO=\sqrt {x^2+a^2}+\sqrt {(b-x)^2+c^2}$
Step 4. Take the derivative to get $s'=\frac{2x}{\sqrt {x^2+a^2}}+\frac{-2(b-x)}{\sqrt {(b-x)^2+c^2}}$. Letting $s'=0$, we have $2x\sqrt {(b-x)^2+c^2}=2(b-x)\sqrt {x^2+a^2}$ which leads to $x^2((b-x)^2+c^2)=(b-x)^2(x^2+a^2)$ or $c^2x^2=a^2(b-x)^2$. Considering the domain for $x$, we have $cx=a(b-x)$ and $x=\frac{ab}{a+c}$
Step 5. To prove $\theta_1=\theta_2$, we need to show that $tan\theta_1=tan\theta_2$ which means that $\frac{x}{a}=\frac{b-x}{c}$ or $cx=a(b-x)$ and this is a result we have shown in step 4.
Step 6. We can use $x=\frac{ab}{a+c}$ and test the signs of $s'$ to show that the corresponding $s(x)$ reaches a minimum. Or we can show $s''\gt0$ as the following:
$s''=2(x^2+a^2)^{-1/2}-\frac{1}{2}(2x)(x^2+a^2)^{-3/2}(2x)+2((b-x)^2+c^2)^{-1/2}-\frac{1}{2}(2x)((b-x)^2+c^2)^{-3/2}(-2(b-x))=2(x^2+a^2)^{-1/2}-2x^2(x^2+a^2)^{-3/2}+2((b-x)^2+c^2)^{-1/2}+2x(b-x)((b-x)^2+c^2)^{-3/2}=2(x^2+a^2)^{-3/2}(x^2+a^2-x^2)+2((b-x)^2+c^2)^{-3/2}((b-x)^2+c^2+2x(b-x))=2a^2(x^2+a^2)^{-3/2}+2((b-x)^2+c^2)^{-3/2}((b-x)^2+c^2+2x(b-x))$
Thus $s''\gt0$ (because each term is greater than zero).
