Edits to “Statement”, “Solution”, “Answer”

Tete edited
revision #20056 parent #20055 ← older newer →
@@ -1,21 +1,21 @@
### Statement
−$13.1.20.$ [Insert the problem statement]
+$13.1.20.$ What shape should the reflecting surface be so that it collects all parallel rays at one point, regardless of the beam width?
### Solution
−![For problem $13.1.20$ |630x840, 31%](../../img/13.1.20/Savchenko.png)
+![For problem $13.1.20$ |630x840, 80%](../../img/13.1.20/Savchenko.png)
We show that a parabolic surface will collect all light rays parallel to its axis of symmetry to its focal point as follows. Suppose that the equation of the parabola in the figure is $y^2=4cx$, so its focal point is at $F(c,0)$. Let $P(a^2/(4c),a)$ be any point on the parabola. (In the figure, $a$ is negative.) Then the slope of the tangent line (shown in blue) at $P$ is
\[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\]
and the slope of the normal line (shown in green) at $P$ is $-a/(2c)$. The slope of $PF$ is
−\[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2)}=\tan(2\angle APB.)\]
+\[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2}=\tan(2\cdot\angle APB.)\]
−In other words, the red lines obey the law of reflection, so an arbitrary light ray parallel to the axis of symmetry of the parabola is collected at its focal point.
+Consequently, $\angle APF=2\cdot\angle APB$; that is, the red lines obey the law of reflection. Therefore, an arbitrary light ray parallel to the axis of symmetry of the parabola is collected at its focal point.
#### Answer
−[Insert a concise answer or boxed result]
+Parabolic surface