Question

A ray of light moving parallel to the \[x-\]axis gets reflected from a parabolic mirror whose equation is ${{\left( y-2 \right)}^{2}}=4\left( x+1 \right)$. After reflection, the ray must pass through the point
(a). $\left( -2,0 \right)$
(b). $\left( -1,2 \right)$
(c). $\left( 0,2 \right)$
(d). $\left( 2,0 \right)$

Answer 1

Hint: The ray of light moving parallel to the \[x-\]axis that gets reflected from a parabolic mirror would pass through the focal point of the parabolic mirror.

Complete step-by-step solution -
It is given in the question that the ray is parallel to the \[x-\]axis. The equation of the parabolic mirror is given as ${{\left( y-2 \right)}^{2}}=4\left( x+1 \right)$.
The plot of the parabolic mirror can be drawn as shown below,


The axis of the parabolic mirror is parallel to the \[x-\]axis and the ray is also parallel to it.
It is known that in a parabolic mirror, the incoming ray of light parallel to the axis is reflected through the focal point.
So, it would mean that, after reflection, the ray of light we have been given in the question would pass through the focal point of the parabolic mirror ${{\left( y-2 \right)}^{2}}=4\left( x+1 \right)$.
We know that the focal point of a parabola of the form ${{y}^{2}}=4ax$ is given as $\left( a,0 \right)$.
No, on comparing the equation ${{\left( y-2 \right)}^{2}}=4\left( x+1 \right)$ with the general equation ${{y}^{2}}=4ax$, we can write the corresponding terms as,
$y=\left( y-2 \right)$, $x=\left( x+1 \right)$ and $a=1$
We know that the coordinates of the focal point for the general equation are $\left( a,0 \right)$. To obtain the focal point of the parabolic mirror, we have to equate both the coordinates as below,
$\left( x+1 \right)=a\Rightarrow \left( x+1 \right)=1\Rightarrow x=0$
$\left( y-2 \right)=0\Rightarrow y=2$
Therefore, the focal point of the parabolic mirror ${{\left( y-2 \right)}^{2}}=4\left( x+1 \right)$ is $\left( 0,2 \right)$.
We have obtained option (c) as the correct answer.

Note: The focal point of a parabola of the form ${{y}^{2}}=4ax$ is given by $\left( a,0 \right)$. There is a chance that you might end up choosing the wrong option by assuming that the focal point for any parabola of the form ${{y}^{2}}=4ax$ would have the y-coordinate as $0$.