# Parabola

#### sbhatnagar

##### Active member
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

Staff member
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
Here are my ideas. Let $P$ be the point of incidence. $P$ is the intersection of the line $y=b$ and the parabola $y^2=4ax$.

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.