📂Geometry

Proof of the Optical Properties of a Parabola

Theorem

For a point $P$ on a parabola, its focus $F$, and the directrix $l$, let the angle between the tangent at $P$ and the line $\overline{PF}$ (which is perpendicular to the directrix $l$) be $\alpha$, and let the angle between the tangent and the line to the focus be $\beta$. Then $\alpha$ and $\beta$ are equal.

Remark

In plain terms, light emanating from the focus of a parabola propagates in a direction perpendicular to the directrix.

Because the rays travel in straight lines, they can convey energy efficiently, which is why parabolic shapes are used in devices such as headlights and satellite-dish antennas.

Proof ¹

$$ y^{2} = 4 a x $$ Given that the parabola is described by the equation above, let the point where the tangent at $P$ meets the $x$-axis be $Q$. For the parameter $t$, the equation of the tangent and the coordinates of $P(t)$ and $Q(t)$ are, respectively, as follows. $$ \begin{align*} y =& {\frac{ 1 }{ t }} x + a t \\ P(t) =& \left( a t^{2} , 2 a t \right) \\ Q(t) =& \left( - a t^{2} , 0 \right) \end{align*} $$ The length of $\overline{QF}$ is given by $$ \left| - a t^{2} - a \right| = a \left( t^{2} + 1 \right) $$ The length of $\overline{FP}$ is given by $$ \begin{align*} & \sqrt{ \left( a t^{2} - a \right)^{2} + \left( 2 a t - 0 \right)^{2} } \\ =& \sqrt{ a^{2} t^{2} - 2 a^{2} t^{2} + a^{2} + 4 a^{2} t^{2} } \\ =& \sqrt{ a^{2} t^{4} + 2 a^{2} t^{2} + a^{2} } \\ =& a \sqrt{ t^{4} + 2 t^{2} + 1 } \\ =& a \left( t^{2} + 1 \right) \end{align*} $$

Since the lengths of the two segments $\overline{QF}$ and $\overline{FP}$ are equal, triangle $\triangle{FPQ}$ is isosceles, and we can conclude $\alpha = \beta$.

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See also

• Optical properties of the parabola
• Optical properties of the ellipse
• Proof of the optical property of the hyperbola