Question

A ray of light is moving along the line $y=x+2,$ gets reflected from a parabolic mirror whose equation is
$y^2=4(x+2).$ The reflection does not happen at the vertex of the parabola. After reflection, the ray does not pass through the focus of the parabola. What will be the equation of the line which contains the reflected ray?

• A. $x-7y-10=0$
• B. $x-7y+26=0$
• C. $x+7y+10=0$
• D. $x+7y-10=0$

Answer 1

The correct option is B $x-7y+26=0$

Let the line $y=x+2$ intersect the parabola at $P$.
Solving the parabola and the line,
$(x+2{)}^2=4(x+2)$
$(x+2)=4$
$x=2$
$y=2+2=4.$

Therefore, the point $P$ is $(2,4).$
Now the equation of the tangent to the parabola
$y^2=4x+8$ at $P(2,4)$ is $y{y}_1=2(x+x_1)+8$
$\Rightarrow 4y=2(x+2)+8$
$\Rightarrow x-2y+6=0$


Let $IP$ be the incident ray, $PM$ be the reflected ray and $PN$ be the normal. We know that the normal is equally inclined with the incident ray and the reflected ray.
We have the slope of $IP=1,$ Slope of tangent at $P=1/2.$ Thus, the slope of normal $PN=-2.$
Let the slope of the reflected ray $PM$ be $m$.
Then, angle between the incident ray $IP$ and normal $PN$ is equal to the angle between the normal $PN$ and reflected ray $PM$.

Thus, using the formula of angle between two lines, we get
$\frac{1+2}{1-2}=\pm \frac{-2-m}{1-2m}$
$\frac{3}{(-1)}=\frac{-2-m}{1-2m}$ or $\frac{3}{(-1)}=-\left(\frac{-2-m}{1-2m}\right)$
$\Rightarrow m=\frac{1}{7}$ or $1$

Now, we cannot take $m=1$ as $m=1$ will give us the original incident ray itself.
Hence, the equation of the reflected ray is
$y-4=\frac{1}{7}(x-2)$
$\Rightarrow 7y-28=x-2$
$\Rightarrow x-7y+26=0$
Hence, the correct option is $\left(a\right).$