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Parabola

sbhatnagar

Active member
Jan 27, 2012
95
A ray of light is coming along the line $y=b$,($b>0$), from the positive direction of the x-axis and strikes a concave mirror whose intersection with the $x-y$ plane is the parabola $y^2 = 4ax$,($a>0$). Find the equation of the reflected ray and show that it passes through the focus of the parabola.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The point on the mirror where the ray strikes it is:

$\displaystyle \left(\frac{b^2}{4a},b \right)$

The law of reflection states that the angle of incidence is equal to the angle of reflection.

Let $\displaystyle x_r=my+k$ represent the path of the reflected ray. We need to show that $\displaystyle (k,0)$ is the focus of the parabola, and is in fact independent of $\displaystyle b$.

The law of reflection gives us:

$\displaystyle \frac{\pi}{2}-\tan^{-1}\left(\frac{dx}{dy} \right)=\tan^{-1}\left(\frac{dx}{dy} \right)+\pi-\tan^{-1}(m)$

Simplifying and using $\displaystyle \frac{dx}{dy}=\frac{y}{2a}$ we have:

$\displaystyle \tan^{-1}(m)-\frac{\pi}{2}=2\tan^{-1}\left(\frac{y}{2a} \right)$

Taking the tangent of both sides, using a co-function identity on the left and a double-angle identity on the right, we have:

$\displaystyle -\frac{1}{m}=\frac{\frac{y}{a}}{1-\left(\frac{y}{2a} \right)^2}$

$\displaystyle m=\frac{\left(\frac{y}{2a} \right)^2-1}{\frac{y}{a}}=\frac{y^2-4a^2}{4ay}$

We may now state, using the point $\displaystyle \left(\frac{b^2}{4a},b \right)$:

$\displaystyle k=\frac{b^2}{4a}-\left(\frac{b^2-4a^2}{4ab} \right)b=\frac{b^2-b^2+4a^2}{4a}=a$

We know the focus of the given parable is at $\displaystyle (a,0)$, thus we have shown the reflected ray will pass through the focus.
 

sbhatnagar

Active member
Jan 27, 2012
95
Here are my ideas. Let $P$ be the point of incidence. $P$ is the intersection of the line $y=b$ and the parabola $y^2=4ax$.

parabola- figure.png

The point $P$ will be $\left( \frac{b^2}{4a},b\right)$.
The equation of tangent $PT$ at $P$ is

$y \cdot b=2a\left( x+\frac{b^2}{4a}\right)$

The slope of this line is $\displaystyle \tan(\theta)=\frac{2a}{b}$(see the diagram)

Let the slope of the reflected ray be $m$.

$$ \begin{aligned} \therefore \ \tan \theta &= \Bigg| \frac{m-\frac{2a}{b}}{1+m\frac{2a}{b}}\Bigg| \\ \frac{2a}{b} &=\Bigg| \frac{m-\frac{2a}{b}}{1+m\frac{2a}{b}}\Bigg|\end{aligned} $$

From here, $\displaystyle m= \frac{4ab}{b^2-4a^2}$

Therefore, the equation of the reflected ray is

$\displaystyle y-b=\frac{4ab}{b^2-4a^2} \left( x-\frac{b^2}{4a}\right)$

This line satisfies the point $(a,0)$, therefore the reflected ray passes through the focus.